Engineering Mathematics-II
W.B. University of Technology, Kolkata
BABU RAM
Formerly Dean, Faculty of Physical Sciences,
Maharshi Dayanand University, Rohtak
Copyright 2011 Dorling Kindersley (India) Pvt. Ltd
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In memory of
my parents
Smt. Manohari Devi and Sri Makhan Lal
This page is intentionally left blank.
Contents
Preface vii
Symbols and Basic Formulae viii
Roadmap to the Syllabus xii
1
Ordinary Differential Equations
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
1.10
1.11
1.12
1.13
1.14
1.15
1.16
1.17
1.18
1.19
1.20 Applications of Linear Differential
Equations 1.67
1.21 Mass-Spring System 1.69
1.22 Simple Pendulum 1.70
1.23 Solution in Series 1.71
1.24 Bessel’s Equation and Bessel’s
Function 1.81
1.25 Fourier–Bessel Expansion of
a Continuous Function 1.88
1.26 Legendre’s Equation and Legendre’s
Polynomial 1.89
1.27 Fourier–Legendre Expansion of a
Function 1.95
1.28 Miscellaneous Examples 1.96
1.1
Definitions and Examples 1.1
Formulation of Differential Equation 1.1
Solution of Differential Equation 1.3
Differential Equations of First Order 1.4
Separable Equations 1.5
Homogeneous Equations 1.8
Equations Reducible to
Homogeneous Form 1.11
Linear Differential Equations 1.12
Equations Reducible to
Linear Differential Equations 1.14
Exact Differential Equation 1.15
The Solution of Exact Differential
Equation 1.16
Equations Reducible to Exact
Equation 1.18
Applications of First Order and
First Degree Equations 1.25
Linear Differential Equations 1.37
Solution of Homogeneous
Linear Differential Equation with
Constant Coefficients 1.39
Complete Solution of Linear Differential
Equation with Constant Coefficients 1.42
Method of Variation of Parameters to
Find Particular Integral 1.51
Differential Equations with
Variable Coefficients 1.54
Simultaneous Linear Differential
Equations with Constant Coefficients 1.65
Exercises
2
Graphs
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
2.10
2.11
2.12
2.13
2.14
1.114
2.1
Definitions and Basic Concepts 2.1
Special Graphs 2.3
Subgraphs 2.6
Isomorphisms of Graphs 2.9
Walks, Paths and Circuits 2.11
Eulerian Paths and Circuits 2.15
Hamiltonian Circuits 2.21
Matrix Representation of Graphs 2.29
Planar Graphs 2.30
Colouring of Graph 2.37
Directed Graphs 2.41
Trees 2.44
Isomorphism of Trees 2.49
Representation of Algebraic Expressions
by Binary Trees 2.52
2.15 Spanning Tree of a Graph 2.55
2.16 Shortest Path Problem 2.57
vi
„ Contents
2.17
2.18
2.19
2.20
Minimal Spanning Tree 2.63
Cut Sets 2.68
Tree Searching 2.72
Transport Networks 2.75
Exercises
3
5.4 Miscellaneous Examples
Exercises
6
3.1
3.1. Improper Integral 3.1
3.2. Convergence of Improper Integral with
Unbounded Integrand 3.1
3.3. Convergence of Improper Integral with
Infinite Limits 3.5
Exercises
4
3.7
Beta and Gamma Functions
4.1
4.1
4.2
4.3
4.4
4.5
Beta Function 4.1
Properties of Beta Function 4.1
Gamma Function 4.5
Properties of Gamma Function 4.5
Relation Between Beta and Gamma
Functions 4.5
4.6 Dirichlet’s and Liouville’s Theorems 4.11
4.7 Miscellaneous Examples 4.13
Exercises 4.14
5
Laplace Transform
5.1
5.1 Definition and Examples of Laplace
Transform 5.1
5.2 Properties of Laplace Transforms 5.8
5.3 Limiting Theorems 5.22
5.26
Inverse Laplace Transform
6.1
6.1 Definition and Examples of Inverse
Laplace Transform 6.1
6.2 Properties of Inverse
Laplace Transform 6.2
6.3 Partial Fractions Method to Find Inverse
Laplace Transform 6.10
6.4 Heaviside’s Expansion Theorem 6.13
6.5 Series Method to Determine Inverse
Laplace Transform 6.14
6.6 Convolution Theorem 6.15
6.7 Complex Inversion Formula 6.20
6.8 Miscellaneous Examples 6.25
2.82
Improper Integrals
5.23
Exercises
7
6.29
Applications of Laplace Transform
7.1
7.1
7.2
7.3
7.4
7.5
7.6
Ordinary Differential Equations 7.1
Simultaneous Differential Equations 7.13
Difference Equations 7.16
Integral Equations 7.21
Integro-Differential Equations 7.24
Solution of Partial Differential
Equation 7.25
7.7 Evaluation of Integrals 7.29
7.8 Miscellaneous Examples 7.31
Exercises
Solved Question Papers
Index
7.36
Q.1
I.1
Preface
All branches of Engineering, Technology and Science require mathematics as a tool for the description
of their contents. Therefore, a thorough knowledge of various topics in mathematics is essential to pursue
courses in Engineering, Technology and Science. The aim of this book is to provide students with sound
mathematics skills and their applications. Although the book is designed primarily for use by engineering
students, it is also suitable for students pursuing bachelor degrees with mathematics as one of the subjects and also for those who prepare for various competitive examinations. The material has been arranged
to ensure the suitability of the book for class use and for individual self study. Accordingly, the contents
of the book have been divided into seven chapters covering the complete syllabus prescribed for B.Tech.
Semester-II of West Bengal University of Technology. A number of examples, figures, tables and exercises
have been provided to enable students to develop problem-solving skills. The language used is simple and
lucid. Suggestions and feedback on this book are welcome.
Acknowledgements
I am extremely grateful to the reviewers for their valuable comments. My family members provided moral
support during the preparation of this book. My son, Aman Kumar, software engineer, Adobe India Ltd,
offered wise comments on some of the contents of the book. I am thankful to Sushma S. Pradeep for
excellently typing the manuscript.
Special thanks are due to Thomas Mathew Rajesh, Anita Yadav, Ravi Vishwakarma, M. E. Sethurajan and
Jennifer Sargunar at Pearson Education for their constructive support.
BABU RAM
Symbols and Basic Formulae
1
a
b
g
Γ
d
Δ
e
ι
q
l
m
n
w
Ω
2
(f) b2 − 4ac < 0 ⇒ the roots are complex
Greek Letters
alpha
beta
gamma
capital gamma
delta
capital delta
epsilon
iota
theta
lambda
mu
nu
omega
capital omega
f
Φ
y
Ψ
x
h
z
c
π
s
Σ
t
r
k
phi
capital phi
psi
capital psi
xi
eta
zeta
chi
pi
sigma
capital sigma
tau
rho
kapha
if b2 − 4ac is a perfect square, the roots
are rational
(g)
3
Properties of Logarithm
(i)
loga 1 = 0, loga 0 = −∞ for a > 1, loga a = 1
loge 2 = 0.6931, loge 10 = 2.3026,
log10 e = 0.4343
(ii) loga p + loga q = loga pq
(iii) loga p + loga q = loga
(iv)
(v)
4
log a p q = q log a p
log a n = log a b ⋅ log b n =
− b ± b − 4 ac
2a
b
the sum of the roots is equal to −
a
c
product of the roots is equal to
a
b2 − 4ac = 0 ⇒ the roots are equal
b2 − 4ac > 0 ⇒ the roots are real
and distinct
5
Algebraic Signs of Trigonometrical Ratios
(a) First quadrant: All trigonometric ratios are
positive
(b) Second quadrant: sin q and cosec q are positive,
all others negative
(c) Third quadrant: tan q and cot q are positive, all
others negative
(d) Fouth quadrant: cos q and sec q are positive, all
others negative
6
Commonly Used Values of Trigonometrical Ratios
sin
π
2
2
(a)
(b)
(c)
(d)
(e)
its roots are given by
180°
π
(ii) 1° = 0.0174 radian
Algebraic Formulae
(vi) If ax2 + bx + c = 0 is quadratic, then
log b n
log b a
Angles Relations
(i) 1 radian =
(i) Arithmetic progression a, a + d, a + 2d,
nth term Tn = a + (n − 1) d
n
Sum of n terms = [2a + ( n − 1)d ]
2
(ii) Geometrical progression: a, ar, ar2,
nth term Tn = ar n−1
a(1 − r n )
Sum of n terms =
1− r
(iii) Arithmetic mean of two numbers a and b is
1
( a + b)
2
(iv) Geometric mean of two numbers a and b
is ab
2ab
(v) Harmonic mean of two numbers a and b is
a+b
p
q
cosec
= 1, cos
π
2
π
2
= 1, sec
= 0, tan
π
2
π
2
=∞
= ∞, cos
π
2
=0
1
π
3
π
1
= , cos =
, tan =
2
6
2
6
3
π
π
π
2
cosec = 2, sec =
, cot = 3
6
6
6
3
sin
sin
π
6
π
3
=
π 1
π
3
, cos = , tan = 3
2
3 2
3
Symbols and Basic Formulae „
2
π
π
1
, sec = 2, cot =
3
3
3
3
3
π
π
π
1
1
sin =
, cos =
, tan = 1
4
4
4
2
2
cosec
cosec
7
π
π
4
=
= 2, sec
π
4
= 2, cot
π
4
(m)
=1
(n)
(o)
Trigonometric Ratios of Allied Angles
(a)
sin( −θ ) = − sin θ , cos( −θ ) = cos θ
tan( −θ ) = − tan θ
cosec( −θ ) = − cosec θ , sec( −θ ) = sec θ
cot( −θ ) = − cot θ
(p)
(q)
(r)
(b) Any trigonometric ratio of
(n.90 ± θ ) =
(s)
⎧± same trigonometric ratio of θ
⎪
⎨when n is even
⎪± co-ratio of θ when n is odd
⎩
For example: sin(4620) = sin[90°(52) − 60°]
(t)
9
3
.
2
Similarly, cosec(270° − θ ) = cosec(90°(3) − θ )
= − sec θ .
= sin( −60°) = − sin 60° = −
8
Transformations of Products and Sums
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
sin( A + B ) = sin A cos B + cos A sin B
sin( A − B ) = sin A cos B − cos A sin B
cos( A + B ) = cos A cos B − sin A sin B
cos( A − B ) = cos A cos B + sin A sin B
tan A + tan B
tan( A + B ) =
1 − tan A tan B
tan A − tan B
tan( A − B) =
1 + tan A tan B
2 tan A
sin 2 A = 2sin A cos A =
1 + tan 2 A
cos 2 A = cos 2 A − sin 2 A = 1 − 2sin 2 A
1 − tan 2 A
1 + tan 2 A
sin 2 A
2 tan A
(i) tan 2 A =
=
cos 2 A 1 − tan 2 A
( j) sin 3 A = 3sin A − 4sin 3 A
= 2cos 2 A − 1 =
(k) cos3 A = 4 cos3 A − 3cos A
3tan A − tan 3 A
1 − 3tan 2 A
A+ B
A− B
sin A + sin B = 2sin
cos
2
2
A+ B
A− B
sin
sin A − sin B = 2cos
2
2
A+ B
A− B
cos A + cos B = 2cos
cos
2
2
A+ B
B−A
cos A − cos B = 2sin
sin
2
2
1
sin A cos B = [sin( A + B) + sin( A − B)]
2
1
cos A sin B = [sin( A + B) − sin( A − B)]
2
1
cos A cos B = [cos( A + B ) + cos( A − B )]
2
1
sin A sin B = [cos( A − B) − cos( A + B )]
2
(l) tan 3 A =
Expressions for sin A2 ; cos A2 and tan
(a) sin 2A = ±
1 − cos A
2
(b) cos 2A = ±
1 + cos A
2
(c) tan 2A = ±
1 − cos A
1 + cos A
A
2
(d) sin 2A + cos 2A = ± 1 + sin A
(e) sin 2A − cos 2A = ± 1 − sin A
10
Relations Between Sides and Angles of a Triangle
a
b
c
(sine formulae)
=
=
sin A sin B sin C
b2 + c2 − a2 ⎫
(b) cos A =
⎪
2bc
⎪
2
2
2
c +a −b ⎪
cos B =
⎬ cosine formulae
2ca
⎪
a2 + b2 − c2 ⎪
cos C =
⎪
2ab
⎭
cos
cos
=
+
a
b
C
c
B
⎫
(c)
⎪
b = c cos A + a cos C ⎬ Projection formulae.
c = a cos B + b cos A⎪⎭
(a)
ix
x
„ Symbols and Basic Formulae
11
d
1
(cosec −1 x ) = −
dx
x x2 − 1
d
(s)
(sinh x ) = cosh x
dx
d
(t)
(cosh x ) = sinh x
dx
(u) D n (uv ) = D nu + nc1D n −1uDv + nc2 D n − 2uD 2 v
Permutations and Combinations Formulae
(r)
n!
,
( n − r )!
n!
n
= nCn − r ,
Cr =
r !( n − r )!
n
Pr =
n
C0 = nCn = 1
+ + n Cr D n − r uD r v + + n CnuD n v
12
(Leibnitz’s Formula)
Differentiation Formulae
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
( j)
(k)
(l)
(m)
d
(sin x ) = cos x
dx
d
(cos x ) = − sin x
dx
d
(tan x ) = sec2 x
dx
d
(cot x ) = − cosec2 x
dx
d
(sec x ) = sec x tan x
dx
d
(cosec x ) = − cosec x cot x
dx
d x
(e ) = e x
dx
d x
( a ) = a 2 log e a
dx
1
d
(log a x ) =
dx
x log a
1
d
(log e x ) =
dx
x
d
( ax + b) n = na( ax + b) n −1
dx
dn
( ax + b) m = m( m − 1)( m − 2)
dx n
…( m − n + 1)( ax + b) m − n
1
d
(sin −1 x ) =
dx
1 − x2
1
d
(cos −1 x ) = −
dx
1 − x2
1
d
(o)
(tan −1 x ) =
1 + x2
dx
1
d
(p)
(cot −1 x ) = −
1 + x2
dx
1
d
(q)
(sec −1 x ) =
dx
x x2 − 1
(n)
13
Integration Formulae
(a)
(b)
(c)
(d)
(e)
(f )
(g)
(h)
(i)
( j)
(k)
(l)
(m)
(n)
(o)
(p)
∫ sin x dx = − cos x
∫ cos x dx = sin x
∫ tan x dx = − log cos x
∫ cot x dx = logsin x
∫ sec x dx = log(sec x + tan x)
∫ cosec x dx = log(cosec x − cot x)
∫ sec x dx = tan x
∫ cosec x dx = − cot x
∫ e dx = e
2
2
x
x
ax
∫ a dx = log
x
e
a
1
∫ x dx = loge x
x n +1
n
∫ x dx = n + 1 , n ≠ −1
1 −1 x
dx
∫ a2 + x 2 = a tan a
1
dx
a+ x
∫ a2 − x 2 = 2a loge a − x
1
dx
x−a
∫ x 2 − a2 = 2a loge x + a
dx
x
= sin −1
a
a2 − x 2
(q)
∫
(r)
∫
(s)
∫
dx
a2 + x 2
dx
x −a
2
2
= sinh −1
x
a
= cosh −1
x
a
a 2 + x 2 dx =
x a2 + x 2 a2
x
+ sinh −1
2
2
a
Symbols and Basic Formulae „
(t)
∫
x x 2 − a2 a2
x
− cosh −1
2
2
a
x 2 − a 2 dx =
x a −x
a
x
+ sin −1
2
2
a
e ax
(v) ∫ e ax sin bx dx = 2
( a sin bx − b cos bx )
a + b2
e ax
(w) ∫ e ax cos bx dx = 2
( a cos bx + b sin bx )
a + b2
(u)
(x)
2
∫
∫
2
2
a 2 − x 2 dx =
π
2
0
π
sin n x dx = ∫ 2 cos n x dx
∫
π
2
0
Beta and Gamma Functions
1
(a) b ( m, n) = ∫ x m−1 (1 − x )n−1 dx converges for m, n > 0
0
∞
(b) Γ ( n) = ∫ e − x x n−1dx converges for n > 0
0
(c) Γ ( n + 1) = n Γ ( n) and Γ ( n + 1) = n !
if n is positive integer
⎛ 1⎞
(d) Γ (1) = 1 = Γ (2) and Γ ⎜ ⎟ = π
⎝ 2⎠
(e) b (m, n) =
0
⎧ ( n − 1)( n − 3)( n − 5)……
if n is odd
⎪ n( n − 2)( n − 4)…
⎪
=⎨
⎪ ( n − 1)( n − 3)( n − 5)… π if n is even
⎩⎪ n( n − 2)( n − 4)… 2
(y)
14
(f )
π
2
∫
0
Γ ( m)Γ ( n)
Γ ( m + n)
sin p x cos q x dx =
1 ⎛ p + 1 q + 1⎞
b⎜
,
⎟
2 ⎝ 2
2 ⎠
⎛ p + 1⎞ ⎛ q + 1⎞
Γ⎜
Γ
⎝ 2 ⎟⎠ ⎜⎝ 2 ⎟⎠
=
⎛ p + q + 2⎞
2Γ ⎜
⎟⎠
⎝
2
sin m x cos n x dx
⎧ ( m − 1)( m − 3)…( n − 1)( n − 3)…
⎪ ( m + n)( m + n − 2)( m + n − 4)…
⎪
⎪ if m and n are not simultaneously even
=⎨
⎪ ( m − 1)( m − 3)…( n − 1)( n − 3)… π
⎪ ( m + n)( m + n − 2)( m + n − 4)… 2
⎪
⎩ if both m and n are even
xi
(g)
∫
π
2
0
tan θ dθ = ∫
π
2
0
1 ⎛ 1⎞ ⎛ 3⎞
= Γ⎜ ⎟ Γ⎜ ⎟
2 ⎝ 4⎠ ⎝ 4⎠
⎛ 3⎞ ⎛ 1⎞
Γ⎜ ⎟ Γ⎜ ⎟
⎝ 4⎠ ⎝ 4⎠
sin θ cos θ dθ =
2 Γ (1)
1
2
−1
2
Roadmap to the Syllabus
W.B. UNIVERSITY OF TECHNOLOGY, KOLKATA
MODULE I
Ordinary differential equations (ODE) — First order and first degree: Exact equations, Necessary and
sufficient condition of exactness of a first order and first degree ODE (statement only), Rules for finding
Integrating factors, Linear equation, Bernoulli’s equation. General solution of ODE of first order and higher
degree (different forms with special reference to Clairaut’s equation).
MODULE II
ODE — Higher order and first degree: General linear ODE of order two with constant coefficients, C.F. and
P.I., D-operator methods for finding P.I., Method of variation of parameters, Cauchy-Euler equations, Solution of simultaneous linear differential equations.
)
REFER
Chapter 1
MODULE III
Basics of Graph Theory: Graphs, Digraphs, Weighted graph, Connected and disconnected graphs, Complement of a graph, Regular graph, Complete graph, Subgraph, Walks, Paths, Circuits, Euler Graph, Cut sets
and cut vertices, Matrix representation of a graph, Adjacency and incidence matrices of a graph, Graph
isomorphism, Bipartite graph.
MODULE IV
Tree: Definition and properties, Binary tree, Spanning tree of a graph, Minimal spanning tree, properties of
trees, Algorithms: Dijkstra’s Algorithm for shortest path problem, Determination of minimal spanning tree
using DFS, BFS, Kruskal’s and Prim’s algorithms.
)
REFER
Chapter 2
MODULE V
Improper Integral: Basic ideas of improper integrals, working knowledge of Beta and Gamma functions
(convergence to be assumed) and their interrelations. Laplace Transform (LT): Definition and existence
of LT, LT of elementary functions, First and second shifting properties, Change of scale property; LT of
f (t) t, LT of ( ) n t f t, LT of derivatives of f (t), LT of ∫ f (u)du. Evaluation of improper integrals using LT, LT
of periodic and step functions, Inverse – LT: Definition and its properties; Convolution Theorem (statement
only) and its application to the evaluation of inverse LT, Solution of linear ODE with constant coefficients
(initial value problem) using LT.
)
REFER
Chapters 3–7
1
Ordinary Differential Equations
Differential equations play an important role in
engineering and science. Many physical laws and
relations appear in the form of differential equations. For example, the current I in an LCR circuit
is described by the differential equation LI 00 þ RI þ
1
C I ¼ E, which is derived from Kirchhoff’s laws.
Similarly, the displacement y of a vibrating mass m
on a spring is described by the equation my 00 þ ky ¼ 0.
The study of differential equations involves formation
of differential equations, the solutions of differential
equations, and the physical interpretation of the
solution in terms of the given problem.
1.1
DEFINITIONS AND EXAMPLES
Definition 1.1. A differential equation is an equation
which involves derivatives.
For example,
2
2
2
¼0
(a) ddt2x þ n2 x ¼ 0, (b) ddx2y þ dy
dx
(c)
@2
@x2
þ @@y2 ¼ 0
2
and
(d)
dy
dx
¼xþ1
are differential equations.
We note that a differential equation may have the
variables present only in the derivatives. For example
in (b), the variables are present only in derivatives.
Moreover, a differential equation may have more
than one dependent variable. For example,
d dł
þ
¼þł
dt
dt
has two dependent variables and ł and one
independent variable t.
Definition 1.2. A differential equation involving
derivatives with respect to a single independent
variable is called an ordinary differential equation.
For example,
d2y
þ 3y ¼ 0
dx2
is an ordinary differential equation.
Definition 1.3. A differential equation involving partial
derivatives with respect to two or more independent
variables is called a partial differential equation.
For example,
2
@2u
@u
@2u
2@ u
¼
k
¼
a
and
@t2
@x2
@t
@x2
are partial differential equations.
Definition 1.4. The order of the highest derivative
appearing in a differential equation free from radicals is called the order of that differential equation.
For example the, order of the differential equation
y 00 þ4y ¼ 0 is two, the order of the differential
2
2
equation @@t2u ¼ a2 @@xu2 is two, and order of the difx
ferential equation y ¼ x dy
dx þ dy=dx is one.
Definition 1.5. The degree or exponent of the highest
derivative appearing in a differential equation free
from radicals and fractions is called the degree of
the differential equation.
For example, the degree of the differential equation
2
dy
dy
þ1
y ¼x
dx
dx
is two. Similarly, the degree of the differential equation
3 2=3
d y
dy
¼1þ2
3
dx
dx
is two because the given equation can be written as
3 2 d y
dy 3
¼ 1þ2
:
dx3
dx
1.2
FORMULATION OF DIFFERENTIAL EQUATION
The derivation of differential equations from physical
or other problems is called modelling. The modelling
involves the successive differentiations and elimination of parameters present in the given sytem.
1.2
n
Engineering Mathematics-II
EXAMPLE 1.1
Form the differential equation for \free fall" of a
stone dropped from the height y under the action of
force due to gravity g.
Solution. We know that equation of motion of the
free fall is
1
y ¼ gt2 :
2
Differentiating with respect to t, we get
dy
¼ gt:
dt
Differentiating once more, we get
d2y
¼ g;
dt2
which is the desired differential equation representing the free fall of a stone.
EXAMPLE 1.2
Form the differential equation of simple harmonic
motion given by x ¼ A cos (vt þ ), where A
and are constants.
Solution. To get the required differential equation, we
have to differentiate the given relation and eliminate
the constants A and . Differentiating twice, we get
dx
¼ Av sinðvt þ Þ;
dt
d2x
¼ Av2 cosðvt þ Þ ¼ v2 x:
ð1Þ
dt2
Hence
d2x
þ v2 x ¼ 0
dt2
the lowest point of the curve and let the length of
the arc OQ be s.
y
P(x, y)
y
Q(x, u)
u
0
x
Figure 1.1
By Law of Conservation of energy, we have
potential energy at P + kinetic energy at P
¼ potential energy at Q + kinetic energy at Q;
and so
2
1
ds
mgy þ 0 ¼ mgu þ m
:
2
dt
Hence
2
ds
¼ 2gðy uÞ
dt
is the required differential equation.
If the duration T0 of descent is independent of
the starting point, the solution of this differential
equation comes out to be the equation of a cycloid.
Thus the shape of the curve is a cycloid. This curve
is called Tautochrone Curve.
is the differential equation governing simple harmonic motion. Equation (1) shows that acceleration
varies as the distance from the origin.
EXAMPLE 1.4
Derive the differential equation governing a massspring system.
EXAMPLE 1.3
Find the differential equation governing the motion
of a particle of mass m sliding down a frictionless
curve.
Solution. Let m be the mass suspended by a
spring that is rigidly supported from one end (see
Figure 1.2). Let
Solution. Velocity of the particle at the starting
point P(x, y) is zero since it starts from rest. Let
(x, u) be some intermediate point during the
motion (see Figure 1.1). Let the origin O be
(i) rest position be denoted by x ¼ 0, downward displacement by x > 0 and upward
displacement be denoted by x < 0.
(ii) k > 0 be spring constant and a > 0 be
damping constant.
Ordinary Differential Equations
(iii) a dx
dt be the damping force due to medium
(damping force is proportional to the
velocity).
and then (3) yields
x h ¼ ðy kÞ
(iv) f (t) be the external impressed force on m.
dy
¼
dx
dy
dx
h
1þ
n
1.3
dy2 i
dx
d 2 y=dx2
:
Substituting the values of x – h and y – k in (2),
we get
"
2 2
2 #3
dy
2 d y
¼a
;
1þ
dx
dx2
x <0
x ⫽ 0 (rest position)
0
which is the required differential equation. It
follows that
h
2 i3=2
1 þ dy
dx
¼ a;
d 2 y=dx2
x >0
f(t)
Figure 1.2
Then, by Newton’s second law of motion, the
2
sum of force acting on m is m ddt2x and so
d2x
dx
m 2 ¼ kx a þ f ðtÞ;
dt
dt
that is,
d 2 x a dx k
þ
þ x ¼ f ðtÞ;
dt2 m dt m
and so the radius of curvature of a circle at any point
is constant.
EXAMPLE 1.6
Form the differential equation from the equation
xy ¼ Aex þ B e–x.
Solution. Differentiating twice, we get
x
which is the required differential equation governing
the system.
EXAMPLE 1.5
Find the differential equation of all circles of radius
a and centre (h, k).
x
ð2Þ
Differentiating twice, we get
dy
x h þ ðy kÞ ¼ 0;
dx
d2y
dy 2
1 þ ðy kÞ 2 þ
¼0
dx
dx
x
From (4), we have
2
1 þ dy
y k ¼ 2 dx 2
d y=dx
ð4Þ
d2y
dy
þ 2 xy ¼ 0
2
dx
dx
is the required differential equation.
1.3
ð3Þ
d 2 y dy dy
þ þ ¼ Aex þ B ex ¼ xy:
dx2 dx dx
Hence
Solution. We know that the equation of the circle
with radius a and centre (h, k) is
ðx hÞ2 þ ðy kÞ2 ¼ a2
dy
þ y ¼ Aex B ex
dx
SOLUTION OF DIFFERENTIAL EQUATION
Definition 1.6. A solution of a differential equation is
a functional relation between the variables involved
such that this relation and the derivatives obtained
from it satisfy the given differential equation.
For example, x2 þ 4y ¼ 0 is a solution of the differential equation
2
dy
dy
þx y ¼ 0:
ð5Þ
dx
dx
1.4
n
Engineering Mathematics-II
In fact, differentiating x2 þ 4y ¼ 0, we get
dy
2x þ 4 ¼ 0
dx
dy
x
and so dx ¼ 2. Hence
2
x
dy
dy
x2
þx y ¼ þ x y
4
dx
dx
2
2
x2 x2
x
¼ 0:
¼ 4
2
4
Hence x2 þ 4y ¼ 0 is a solution of (5).
Definition 1.7. A solution of a differential equation in
which the number of arbitrary constants is equal to the
order of the differential equation is called the general
(or complete) solution of the differential equation.
For example xy ¼ Aex þ Be–x is the general solution of
d2y
dy
x 2 þ 2 xy ¼ 0:
dx
dx
Definition 1.8. A solution obtained from the general
solution of a differential equation by giving particular values to the arbitrary constants is called a
particular solution of that differential equation.
Definition 1.9. A problem involving a differential
equation and one or more supplementary conditions,
relating to one value of the independent variable,
which the solution of the given differential equation
must satisfy is called an initial-value problem.
For example,
d2y
þ y ¼ cos 2t;
dt2
yð0Þ ¼ 1; y0 ð0Þ ¼ 2
is an initial-value problem. Similarly, the problem
d2y
þ y ¼ 0;
dx2
yð1Þ ¼ 3; y0 ð1Þ ¼ 4
is also an initial-value problem.
Definition 1.10. A problem involving a differential
equation and one or more supplementary conditions,
relating to more than one values of the independent
variable, which the solution of the differential
equation must satisfy, is called a boundary-value
problem.
For example,
d2y
þy¼0
dx2 yð0Þ ¼ 2; y
¼4
2
is a boundary value problem.
1.4
DIFFERENTIAL EQUATIONS OF FIRST ORDER
We consider first the differential equations of first
order. Let D be a open connected set in R2 and let
f : D ! R be continuous. We discuss the problem of
determining solution in D of the first order differential equation
dy
¼ f ðx; yÞ:
dx
Definition 1.11. A real-valued function f: D ! R
defined on the connected open set D in R2 is said to
satisfy a Lipschitz condition in y on D with
Lipschitz constant M if and only if
j f ðx; y2 Þ f ðx; y1 Þj Mj y2 y1j
for all (x, y1) and (x, y2) 2 D.
Regarding existence of solutions of first order
differential equation, we have the following
theorems.
Theorem 1.1. (Picard’s Existence and Uniqueness
Theorem). Let f: D ! R be continuous on open
connected set D in R2 and satisfy Lipschitz condition
in y on D. Then for every (x0, y0) 2 D, the initial-value
problem dy
dx ¼ f (x, y) has a solution passing through
(x0, y0).
The solution obtained in Theorem 1.1 is unique. The
Lipschitz condition in the hypothesis of Picard’s
theorem cannot be dropped because continuity of f
without this condition will not yield unique solution. For example, consider the equation
dy
¼ y2=3 ; yð0Þ ¼ 0:
dx
Clearly 1(x) ¼ 0 is a solution to this equation.
Further substituting y ¼ sin3h, we have dy ¼ 3 sin2h
cos h dh and so
3 sin2 h cos h dh
¼ sin2 h;
dx
which yields 3 cos h dh ¼ dx and x ¼ 3 sin h. Thus
x3
x3 ¼ 27 sin3h ¼ 27y and hence y ¼ 27
is also a
Ordinary Differential Equations
solution. Therefore, the given initial-value problem
does not have unique solution. The reason is that it
does not satisfy Lipschitz condition.
Solution. We have
n
dy
ex
¼ y
dx e
and so
Theorem 1.2. (Peano’s Existence Theorem). Let f be a
continuous real-valued function on a non-empty open
subset D of the Euclidean space R2 and let (x0, y0) 2 D.
Then there is a positive real number a such that the
first order differential equations dy
dx ¼ f (x, y) has a
solution in the interval [x0, x0 þ a] which satisfies
the boundary condition (x0) ¼ y0.
Clearly, Peano’s theorem is merely an existence theorem and not a uniqueness theorem.
We now consider certain basic types of first
order differential equations for which an exact solution may be obtained by definite procedures. The
most important of these types are separable equations, homogeneous equations, exact equations,
and linear equations. The corresponding methods
of solution involve various devices. We take these
types of differential equations one by one.
1.5
SEPARABLE EQUATIONS
ey dy ¼ ex dx:
Integrating both sides
Z
Z
y
e dy ¼ ex dx þ C or ey ¼ ex þ C:
Using initial condition y (1) ¼ 1, we get
e1 ¼ e þ C
2
and so C ¼ 1þe
.
e
Thus the solution is
ey ¼ ex ey ¼ e
ð6Þ
is separable if f may be expressed as
ðx; yÞ 2 D
dy
¼ MðxÞ
dx
ð7Þ
ð8Þ
The equation (8) is solved by integrating with
respect to x. Thus the solution is
Z
Z
N ð yÞdy ¼ MðxÞdx þ C:
EXAMPLE 1.7
Solve
dy
¼ exþy ; yð1Þ ¼ 1: Find yð1Þ:
dx
or
y ¼ 1:
EXAMPLE 1.8
The function M(x) and N(y) are real-valued functions of x and y, respectively. Thus (6) becomes
N ð yÞ
1 þ e2
e
2
1 1þe
¼ e;
¼ e
e
yð1Þ ¼ e1 The first order differential equation
MðxÞ
;
f ðx; yÞ ¼
N ð yÞ
1 þ e2
:
e
Hence y(1) is given by
that is
dy
¼ f ðx; yÞ
dx
1.5
2
Solve dy
dx ¼ (4x þ y þ 1) , y(0) ¼ 1.
Solution. Substituting 4x þ y þ 1 ¼ t, we get
dy dt
¼ 4
dx dx
Hence, the given equation reduces to
dt
4 ¼ t2
dx
or
dt
¼ t2 þ 4
dx
or
dt
¼ dx:
2
t þ4
Integrating both sides, we have
Z
Z
dt
dt ¼ dx þ C
t2 þ 4
1.6
n
Engineering Mathematics-II
or
1
t
tan1 ¼ x þ C
2
2
or
1
4x þ y þ 1
tan1
¼xþC
2
2
or
4x þ y þ 1 ¼ 2 tan 2ðx þ CÞ
Using given initial conditions x ¼ 0, y ¼ 1, we get
1
tan1 1 ¼ C
2
which gives C ¼ 8. Hence the solution is
4x þ y þ 1 ¼ 2 tanð2x þ =4Þ:
EXAMPLE 1.9
Solve
Solution. Dividing the given equation throughout by
(1 þ x2) (1 þ y2), we get
x
y
dx þ
dy ¼ 0:
1 þ x2
1 þ y2
Integrating both sides, we have
Z
Z
x
y
dx þ
dy ¼ C
1 þ x2
1 þ y2
or
Z
Z
1
2x
1
2y
dx
þ
dy ¼ C
2
2 1þx
2 1 þ y2
or
1
1
logð1 þ x2 Þ þ logð1 þ y2 Þ ¼ C
2
2
or
1
logð1 þ x2 Þð1 þ y2 Þ ¼ C
2
or
logð1 þ x2 Þð1 þ y2 Þ ¼ 2C ¼ log C
dy
dy
:
y x ¼ a y2 þ
dx
dx
Solution. The given equation can be written as
dy
ða þ xÞ ¼ yð1 ayÞ
dx
or
dy
dx
¼
:
yð1 ayÞ a þ x
Integrating both sides, we have
log y logð1 ayÞ ¼ logða þ xÞ þ C
or
log
y
¼ C:
ða þ xÞ ð1 ayÞ
Hence
y ¼ Kða þ xÞð1 ayÞ; K constant
Hence
ð1 þ x2 Þð1 þ y2 Þ ¼ K ðconstantÞ
is the required solution.
EXAMPLE 1.11
Solve
dy
¼ exy þ x2 ey :
dx
Solution. We have
dy ex x2
¼ þ
dx ey ey
or
dy
ey ¼ ex þ x2
dx
or
ey dy ¼ ðex þ x2 Þdx ¼ ex dx þ x2 dx:
Integrating both sides, we get
x3
ey ¼ ex þ þ C ðconstantÞ:
3
is the general solution.
EXAMPLE 1.10
Solve
xð1 þ y2 Þdx þ yð1 þ x2 Þdy ¼ 0:
EXAMPLE 1.12
Solve
16y
dy
þ 9x ¼ 0:
dx
Ordinary Differential Equations
n
1.7
Solution. We are given that
dy
16y þ 9x ¼ 0
dx
Solution. We have
or
dz
¼ 1 – 2
Substituting z ¼ x – 2y, we get dx
The equation becomes
dz
ð2z þ 5Þ ¼ 4z þ 11
dx
or
dz
ð4z þ 10Þ ¼ 2ð4z þ 11Þ
dx
or
4z þ 10
dz ¼ 2 dx
4z þ 11
or
4z þ 11 1
dz ¼ 2dx
4z þ 11
or
1
dz ¼ 2dx
1
4z þ 11
16y dy ¼ 9x dx:
Integrating both sides, we have
16
y2
x2
¼ 9 þ C
2
2
or
x2 y2
þ ¼ K ðconstantÞ;
16 9
which is the required solution and represents a
family of ellipses.
EXAMPLE 1.13
Solve
dy
¼ ðx þ yÞ2 :
dx
dz
Solution. Substituting z ¼ x þ y, we get dx
¼ 1 þ dy
dx.
Therefore, the given equation reduces to
dz
1 ¼ z2
dx
or
dz
¼ 1 þ z2
dx
or
dz
¼ dx:
1 þ z2
Integrating, we get
tan1 z ¼ x þ C:
5 dy
2ðx 2y þ Þ ¼ x 2y þ 3:
2 dx
Integrating both sides, we have
1
z log j4z þ 11j ¼ 2x þ C:
4
Putting back the value of z, we have
4x þ 8y þ log j4x 8y þ 11j ¼ C:
EXAMPLE 1.15
Solve (x þ 2y) (dx – dy) ¼ dx þ dy.
Solution. We have
ðx þ 2yÞðdx dyÞ ¼ dx þ dy
Putting back the value of z, we get
tan1 ðx þ yÞ ¼ x þ C:
dy
dx.
or
ðx þ 2y 1Þdx ¼ ðx þ 2y þ 1Þdy
Hence
x þ y ¼ tanðx þ CÞ:
EXAMPLE 1.14
Solve
ð2x 4y þ 5Þ
dy
¼ x 2y þ 3:
dx
or
dy x þ 2y 1
¼
:
dx x þ 2y þ 1
dz
Substituting x þ 2y ¼ z, we have 1 þ 2dy
dx ¼ dx and so
dy 1 dz
¼
1 :
dx 2 dx
1.8
n
Engineering Mathematics-II
Hence the above equation becomes
1 dz
z1
1 ¼
2 dx
zþ1
or
dz 3z 1
¼
:
dx
zþ1
Separating the variables, we have
ðz þ 1Þ dz
¼ dx
3z 1
or
or
EXAMPLE 1.16
Solve
1 3z 1 þ 4
dz ¼ dx
3
3z 1
ðx2 þ y2 Þdx 2xy dy ¼ 0:
Solution. We have
1
4
1þ
dz ¼ dx:
3
3z 1
Integrating both sides, we get
Z Z
1
4
1þ
dz ¼ dx þ C
3
3z 1
or
where f (x, y) and (x, y) are homogeneous functions of the same degree in x and y is called an
homogeneous equation.
A homogeneous differential equation can be
dv
solved by substituting y ¼ vx. Then dy
dx ¼ v þ x dx.
dy
Putting the value of y and dx in the given equation,
we get a differential equation in which variables can
be separated. Integration then yields the solution in
terms of v, which we replace with yx.
1
4
z þ logð3z 1Þ ¼ x þ C
3
3
or
3z þ 4 logð3z 1Þ ¼ 9x þ k ðconstantÞ
ðx2 þ y2 Þdx 2xy dy ¼ 0
and so
dy x2 þ y2
¼
2xy
dx
This equation is homogeneous in x and y. So, put
dv
y ¼ vx. We have dy
dx ¼ v þ x dx. Hence (9) becomes
vþx
or
1.6
HOMOGENEOUS EQUATIONS
2v dv
dx
¼
1 v2
x
or
dv ¼
dx
:
x
Integrating both sides, we get
logð1 vÞ logð1 þ vÞ ¼ log x þ C
or
logð1 v2 Þ ¼ log x þ C
Definition 1.13. A differential equation of the form
dy f ðx; yÞ
¼
;
dx ðx; yÞ
1
1
1v 1þv
th
in which every term is of the n degree, is called a
homogeneous function of degree n.
dv 1 þ v2
1 v2
v¼
¼
2v
2v
dx
or
Definition 1.12. An expression of the form
a0 xn þ a1 xn1 y þ a2 xn2 y2 þ . . . þ an yn ;
dv x2 þ v2 x2 1 þ v2
¼
¼
2vx2
2v
dx
or
x
k
3ðy xÞ þ 2 logð3x þ 6y 1Þ ¼ ¼ K ðconstantÞ:
2
ð9Þ
or
log x þ logð1 v2 Þ ¼ C
Ordinary Differential Equations
or
log xð1 v Þ ¼ C
2
or
y2 ¼C
log x 1 x
or
y2
x 1 2 ¼ C:
x
Hence
x2 y2 ¼ Cx
is the general solution of the given differential
equation.
EXAMPLE 1.17
Solve
x
Thus
x
dv pffiffiffiffiffiffiffiffiffiffiffiffi2ffi
¼ 1þv
dx
and so
dv
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ :
2
x
1þv
Integrating both sides, we get
Z
Z
dv
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
þC
2
x
1þv
or
logðv þ
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1 þ v2 Þ ¼ log x þ log C
¼ log x C:
Hence
which is the required solution.
EXAMPLE 1.18
Solve
ðy2 2xyÞdx ¼ ðx2 2xyÞdy:
Solution. The given differential equation can be
expressed by
dy y2 2xy
:
¼
dx x2 2xy
Clearly it is a homogeneous equation. Put y ¼ vx so
dv
that dy
dx ¼ vþ x dx. Hence
vþx
dy v2 x2 2x2 v v2 2v
¼
¼ 2
:
dx
x 2x2 v
1 2v
or
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
dy y þ x2 þ y2
;
¼
x
dx
which is homogeneous in x and y. Put y ¼ vx so that
dy
dv
dx ¼ v þ xdx. Hence the equation takes the form
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
dv vx þ x2 þ v2 x2
vþx ¼
dx
x ffi
pffiffiffiffiffiffiffiffiffiffiffiffi
¼ v þ 1 þ v2 :
pffiffiffiffiffiffiffiffiffiffiffiffiffi
v þ 1 þ x2 ¼ x C:
1.9
Substituting the value of v, we get
rffiffiffiffiffiffiffiffiffiffiffiffiffi
y
y2
þ 1 þ 2 ¼ x C;
x
x
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
dy
¼ y þ x 2 þ y2 :
dx
Solution. We have
n
x
Thus
dv v2 2v
¼
v:
dx
1 2v
v2 2v v þ 2v2
¼
ð1 2vÞ
2
3v 3v
:
¼
1 2v
1 2v
dx
dv ¼
2
3v 3v
x
or
1 2v
dx
¼ :
2
3ðv vÞ
x
Integrating, we get
1
logðv2 vÞ ¼ log x þ C
3 1
y2 y
¼ log x þ C
log 2 x
3
x
2
1
y xy
¼ log x þ C
log
3
x2
2
y xy
log x3
¼ log C:
x2
Hence
xðy2 xyÞ ¼ C:
1.10
n
Engineering Mathematics-II
EXAMPLE 1.19
Solve
Now, variables separation yields
x
2v
dx
dv ¼ :
2
1 3v
x
dy y2
þ ¼ y:
dx x
Solution. We have
dy
y2 xy y2
x ¼y ¼
x
x
dx
or
dy xy y2
;
¼
x2
dx
which is homogeneous in x and y. Putting y ¼ vx, we
dv
have dy
dx ¼ v þ x dx Hence the equation becomes
vþx
dy xy y
vx v x
¼
¼ v v2
¼
x2
x2
dx
2
2
Integrating both sides, we get
Z
Z
1
6v
dx
dv ¼
þC
3 1 3v2
x
or
1
logð1 3v2 Þ ¼ log x þ C
3
or
logð1 3v2 Þ ¼ log x3 þ 3C
or
log x3 ð1 3v2 Þ ¼ 3C
2 2
or
or
x
dv
¼ v2
dx
x
3
3y2
1 2
x
¼K
or
or
xðx2 3y2 Þ ¼ K:
dv
dx
¼ :
v2
x
Integrating, we get
1
¼ log x þ log C:
v
Putting v ¼ yx, we get
x
¼ log x C:
y
EXAMPLE 1.21
Solve
x
ð1 þ ex=y dx þ ex=y ð1 Þdy ¼ 0:
y
Solution. We have
ex=y ð1 xyÞ
dx
:
¼
dy
1 þ ex=y
Putting x ¼ vy, we have dx
dy ¼ v þ y
above equation reduces to
EXAMPLE 1.20
Solve
ðx2 y2 Þdx ¼ 2xy dy:
Solution. We have
dy x2 y2
¼
2xy
dx
so that the given equation is homogeneous in x and y.
Putting y ¼ vx, the equation takes the form
dv x2 v2 x2 1 v2
:
¼
vþx ¼
2vx2
2v
dx
Therefore,
dv 1 v2
1 v2 2v2 1 3v2
v¼
¼
:
x ¼
2v
2v
2v
dx
vþy
dv
dy
and so the
dv
ev ð1 vÞ ev ðv 1Þ
¼
:
¼
dy
1 þ ev
1 þ ev
Hence
y
dv ev ðv 1Þ
v þ ev
¼
v
¼
1 þ ev
dy
1 þ ev
and so separation of variable gives
1 þ ev
dy
dv ¼ :
v
vþe
y
Integrating both sides, we get
Z
Z
1 þ ev
dy
¼
þC
v
vþe
y
Ordinary Differential Equations
n
1.11
We choose h and k such that
or
logðv þ e Þ ¼ log y þ log k
2h k þ 1 ¼ 0; and
h þ 2k 3 ¼ 0:
k
logðv þ ev Þ ¼ log :
y
Solving these two equations, we get h ¼ 15, k ¼ 75.
Hence
dY 2X Y
¼
;
dX X þ 2Y
v
or
Thus
k
v þ ev ¼ :
y
which is homogeneous in X and Y. So put Y ¼ vX.
Then
dv 2X vX
2v
¼
¼
vþX
dX X þ 2vX 1 þ 2v
But v ¼ xy. Hence
x þ yex=y ¼ k ðconstantÞ:
and so
1.7
EQUATIONS REDUCIBLE TO HOMOGENEOUS
FORM
Equations of the form
dy
ax þ by þ C
¼
dx a0 x þ b0 y þ C 0
can be reduced, by substitution, to the homogeneous
form and then solved. Two cases arise:
X
Now separation of variables yields
1 þ 2v
dX
dv ¼
:
2ðv2 þ v 1Þ
X
Integrating both sides, we get
Z
Z
1
1 þ 2v
dX
dv ¼
þC
2
2 v þv1
X
(i) If aa0 6¼ bb0 , then the substitution x ¼ X þ h
and y ¼ Y þ k, where h and k are suitable
constants, makes the given equation homogeneous in X and Y.
or
(ii) If aa0 ¼ bb0 , then the substitution ax þ by ¼ z
serves our purpose.
or
EXAMPLE 1.22
Solve
dv
2v
2v2 2v þ 2
¼
v¼
:
dX 1 þ 2v
1 þ 2v
1
logðv2 þ v 1Þ ¼ log X þ C
2
logðv2 þ v 1Þ ¼ 2 log X þ C ¼ log X2 þ C
or
log X 2 ðv2 þ v 1Þ ¼ log k
dy 2x y þ 1
¼
:
dx x þ 2y 3
Solution. We observe that the condition
satisfied in the present case. So, we put
x¼X þh
and
or
a
a0
6¼ bb0 is
X 2 ðv2 þ v 1Þ ¼ k
or
X2
y ¼ Y þ k:
Therefore, dx ¼ dX and dy ¼ dY and the given
equation reduces to
dY 2ðX þ hÞ ðY þ kÞ þ 1
¼
dX
X þ h þ 2ðY þ kÞ 3
2X Y þ 2h k þ 1
:
¼
X þ 2Y þ h þ 2k 3
Y2 Y
þ 1
X2 X
¼k
or
Y 2 þ YX X 2 ¼ k
or
7
yþ
5
2 7
1
1 2
þ yþ
¼k
xþ
xþ
5
5
5
1.12
n
Engineering Mathematics-II
or
14 1
2 7
y2 þ xy x2 þ
þ
yþ þ
x¼k
5 5
5 5
or
dz 2z þ 2
4z þ 5
¼
þ1¼
dx 2z þ 3
2z þ 3
or
or
2z þ 3
dz ¼ dx:
4z þ 5
y2 þ xy x2 þ 3y þ x ¼ k:
EXAMPLE 1.23
Solve
Integrating both sides, we have
Z
Z
2z þ 3
dz ¼ dx þ C
4z þ 5
dy 2x þ 3y þ 4
¼
:
dx 4x þ 6y þ 5
or
¼ is satisfied
Solution.
in the present case. Hence put 2x þ 3y ¼ z so that
dz
2 þ 3dy
dx ¼ dx . Hence the given equation reduces to
dz 7z þ 22
¼
dx
2z þ 5
or
2z þ 5
dz ¼ dx
7z þ 22
We note that the condition aa0
b
b0
Integrating both
Z sides, we getZ
2z þ 5
dz ¼ dx þ C
7z þ 22
or
2
9
z logð7z þ 22Þ ¼ x þ C:
7
49
Substituting z ¼ 2x þ 3y, we get
14ð2x þ 3yÞ 9 logð14x þ 21y þ 22Þ ¼ 49x þ C
or
21x 42y þ 9 logð14x þ 21y þ 22Þ ¼ C
EXAMPLE 1.24
Solve
ðx þ 2y þ 1Þ dx ¼ ð2x þ 4y þ 3Þdy:
Solution. We have
dy
x þ 2y þ 1
ax þ by þ C
:
¼
¼
dx 2x þ 4y þ 3 a0 x þ b0 y þ C
We observe that aa0 ¼ bb0 ¼ 12. So we put x þ 2y ¼ z
dz=dx1
dy
dz
. The given
and have 1 þ 2dy
dx ¼ dx. Then dx ¼
2
equation now reduces to
ðdz=dxÞ 1
zþ1
¼
2
2z þ 3
Z
1
1
þ
dz ¼
2 2ð4z þ 5Þ
Z
dx þ C
or
1
1
z þ logð4z þ 5Þ ¼ x þ C
2
8
or
4z þ logð4z þ 5Þ ¼ 4x þ k ðconstantÞ
or
4ðx þ 2yÞ þ logð4x þ 8y þ 5Þ ¼ 8x þ k
or
1.8
4ð2y xÞ þ logð4x þ 8y þ 5Þ ¼ k:
LINEAR DIFFERENTIAL EQUATIONS
Definition 1.14. A differential equation is said to be
linear if the dependent variable and its derivatives
occur in the first degree and are not multiplied
together.
Thus a linear differential equation is of the form
dy
þ Py ¼ Q, where P and Q are functions of x only
dx
or
dx
þ Px ¼ Q, where P and Q are funtions of y only:
dy
A linear differential equation of the first order is
called Leibnitz’s linear equation.
To solve the linear differential equation
dy
þ Py ¼ Q;
ð10Þ
R
dx
Pdx
we multiply both sides by e
and get
R
R
R
dy
Pdx
Pdx
þ Pye
¼ Q e Pdx
ð11Þ
e
dx
n
Ordinary Differential Equations
But
d
ðy e
dx
R
Pdx
Þ¼
dy
e
dx
R
Pdx
þ P ye
R
Solution. The given equation can be expressed as
Pdx
dx
x
tan1 y
¼
þ
dy 1 þ y2
1 þ y2
:
Hence (11) reduces to
R
R
d
ðy e Pdx Þ ¼ Q e Pdx ;
dx
which an integration yields
Z
R
R
y e Pdx ¼ Q e Pdx dx þ C;
as the required solution.
R
Pdx
and so is Leibnitz’s linear equation in x. Comparing
tan1 y
1
with dx
dy þ Px ¼ Q, we get P ¼ 1þy2 and Q ¼ 1þy2 .
Therefore
R 1
R
1
dy
P dy
¼ e 1þy2 ¼ etan y :
I:F: ¼ e
Hence the solution of the differential equation is
xetan
The factor e
is called an integrating factor
(I.F.) of the differential equation.
1
Z
y
¼
Z
¼
EXAMPLE 1.25
Solve
tan1 y tan1 y
e
dy þ C
1 þ y2
t et dt þ C; tan1 y ¼ t
¼ t et et þ C
dy
ðx þ 2y3 Þ ¼ y:
dx
¼ ðtan
1
ðintegration by partsÞ
y 1Þ etan
1
y
þ C:
Hence
1
Solution. The given differential equation can be
written as
dx
y ¼ x þ 2y3
dy
or
dx
y x ¼ 2y3
dy
or
dx 1
x ¼ 2y2 :
dy y
1
Comparing with dx
dy þ Px ¼ Q, we have P ¼ – y and
Q ¼ 2y4. The integrating factor is
R
R 1
1
I:F: ¼ e P dy ¼ e y dy ¼ e log y ¼ elog y ¼ y1 :
Therefore, the solution of the differential equation is
Z
xy1 ¼ ð2y2 Þy1 dy þ C
Z
¼ 2y dy þ C ¼ y2 þ C:
EXAMPLE 1.26
Solve
x ¼ tan1 y 1 þ C e tan
dx
¼ tan1 y x:
dy
y
:
EXAMPLE 1.27
Solve
sin 2x
dy
¼ y þ tan x:
dx
Solution. We have
dy
1
tan x
sin x
1
¼ sec2 x:
y¼
¼
2
dx sin 2x
sin 2x 2 cos x sin x 2
Thus, the given equation is linear in y. Now
R
1
I:F: ¼ e cosec 2x dx ¼ e2 log tan x
¼ elogðtan xÞ
1=2
¼ ðtan xÞ1=2 :
Hence, the solution of the given differential equation is
Z
1
1 2
sec xðtan xÞ1=2 dx þ C
yðtan xÞ2 ¼
2
Z
1
¼
ðsec2 xÞ ðtan xÞ1=2 dx þ C
2
1 ðtan xÞ2þ1
þC
¼
2 12 þ 1
1
ð1 þ y2 Þ
1.13
¼ ðtan xÞ1=2 þ C;
1.14
n
Engineering Mathematics-II
which can be expressed as
pffiffiffiffiffiffiffiffiffiffi
y ¼ tan x þ C tan x:
1.9
EQUATIONS REDUCIBLE TO LINEAR
DIFFERENTIAL EQUATIONS
Definition 1.15. An equation of the form
dy
ð12Þ
þ Py ¼ Qyn ;
dx
where P and Q are functions of x is called Bernoulli’s
equation.
The Bernoulli’s equation can be reduced to
Leibnitz’s differential equation in the following
way:
Divide both sides of (12) by yn to get
dy
yn þ P y1n ¼ Q
ð13Þ
dx
Put y1–n ¼ z to give
dy dz
ð1 nÞyn ¼
dx dx
or
dy
1 dz
:
yn ¼
dx 1 n dx
Hence (13) reduces to
1 dz
þ Pz ¼ Q
1 n dx
or
dz
þ Pð1 nÞz ¼ Qð1 nÞ;
dx
which is Leibnitz’s linear equation in z and can be
solved by finding the appropriate integrating factor.
EXAMPLE 1.28
Solve
dy
þ x sin 2y ¼ x2 cos2 y:
dx
Solution. Dividing throughout by cos2 y, we have
dy 2x sin y cos y
¼ x2
sec2 y þ
dx
cos2 y
or
sec2 y
dy
þ 2x tan y ¼ x3 :
dx
dz
Putting tan y ¼ z, we have sec2 y dy
dx ¼ dx. Hence, the
given equation reduces to
dz
ð14Þ
þ 2xz ¼ x3 ;
dx
which is Leibnitz-equation in z and x. The integrating factor is given by
R
R
2
I:F: ¼ e Pdx ¼ e 2x dx ¼ ex :
Hence solution of the equation (14) is
Z
Z
2
2
2
zex ¼ x3 ex dx þ C ¼ xðx2 ex Þdx þ C
Z
Z
1
1
2 x2
¼
2xðx e Þdx þ C ¼
tet dt þ C; x2 ¼ t
2
2
1
2
¼ ðx2 1Þex þ C
2
Putting back the value of z, we get
1
2
2
tan y ex ¼ ðx2 1Þ ex þ C
2
or
1
2
tan y ¼ ðx2 1Þ þ C ex :
2
EXAMPLE 1.29
Solve
dy
þ y ¼ xy3 :
dx
Solution. Dividing throughout by y3, we get
dy
y3 þ y2 ¼ x:
dx
–2
dz
Put y ¼ z. Then dx ¼ 2y3 dy
dx and, therefore, the
above differential equation reduces to
1 dz
þz¼x
2 dx
or
dz
2z ¼ 2x;
dx
which is Leibnitz’s equation in z. We have
R
I:F: ¼ e 2 dx ¼ e2x :
Therefore, the solution of the equation in z is
Z
1
z e2x ¼ 2x:e2x dx þ C ¼ e2x ð2x þ 1Þ þ C
2
Ordinary Differential Equations
or
1
z ¼ x þ þ C e2x
2
or
1
y2 ¼ x þ þ C e2x :
2
EXAMPLE 1.30
Solve
x
dy
þ y ¼ x3 y6 :
dx
Solution. Dividing throughout by y6, we have
dy y5
¼ x2 :
y6 þ
x
dx
dz
Putting y–5 ¼ z, we have – 5y6 dy
dx ¼ dx and so the
preceeding equation transfers to
1 dz z
þ ¼ x2
5 dx x
1.15
For example, the differential equation y2 dx þ 2xy
dy ¼ 0 is an exact equation because it is the total differential of U(x, y) ¼ xy2. In fact the coefficient of dy is
2
@F
@y (xy ) ¼ 2xy.
The following theorem tells us whether a given
differential equation is exact or not.
Theorem 1.3. A necessary and sufficient condition for
the differential equation Mdx þ N dy ¼ 0 to be exact
is
@M @N
¼
;
@y
@x
where M and N are functions of x and y having
continuous first order derivative at all points in the
rectangular domain.
Proof: (1) Condition is necessary. Suppose that the
differential equation Mdx þ Ndy ¼ 0 is exact. Then
there exists a function U(x, y) such that
M dx þ N dy ¼ d U:
or
But, in term of partial derivatives, we have
dz 5z
¼ 5x2 :
dx x
dU ¼
Now
R1
I:F: ¼ e5 x dx ¼ e5 log x ¼ x5 :
Therefore, the solution the above differential equation in z is
Z
z:x5 ¼
n
@U
@U
dx þ
dy:
@x
@y
Therefore,
M dx þ N dy ¼
5x2 x5 dx þ C
@U
@U
dx þ
dy:
@x
@y
Equating coefficients of dx and dy, we get
or
y6 :x5 ¼ 5
Z
x3 dx þ C ¼
5 2
x þC
2
or
1 ¼ ð10 þ Cx2 Þx3 y5 :
1.10
@U
@x
and
N¼
@M
@2U
¼
@y
@y @x
and
@N
@2U
¼
:
@x @x @y
M¼
@U
:
@y
Now
EXACT DIFFERENTIAL EQUATION
Definition 1.16. A differential equation of the form
Mðx; yÞdx þ N ðx; yÞdy ¼ 0
Since partial derivatives of M and N are continuous,
we have
is called an exact differential equation if there exists
a function U U ðx; yÞ of x and y such that
M(x, y)dx þ N(x, y) dy ¼ d U.
Thus, a differential eqution of the form
Mðx; yÞdx þ N ðx; yÞdy ¼ 0 is an exact differential
equation if Mdx þ Ndy is an exact differential.
Hence
@2U
@2U
¼
:
@y @x @x @y
@M @N
¼
:
@y
@x
1.16
n
Engineering Mathematics-II
(2) Condition is sufficient. Suppose that M and N
satisfy
@M @N
¼
:
@y
@x
Let
1.11
THE SOLUTION OF EXACT DIFFERENTIAL
EQUATION
In the proof of Theorem 1.3, we note that if M dx þ
N dy ¼ 0 is exact, then
Z
M dx þ N dy ¼ dðU þ
Z
M dx ¼ U ;
where y is treated as a constant while integrating M.
Then
Z
@
@U
M dx ¼
@x
@x
or
M¼
@U
:
@x
Therefore,
@M
@2U
¼
@y
@y @x
Therefore,
f ð yÞdyÞ:
Z
dðU þ
f ð yÞ dyÞ ¼ 0
or
dU þ f ð yÞdy ¼ 0
Integrating, we get the required solution as
Z
U þ f ð yÞ dy ¼ CðconstantÞ
or
Z
Z
M dx þ
f ðyÞ dy ¼ C
y constant
and so
@M
@2U
@N
¼
¼
:
@y
@y @x @x
Also, by continuity of partial derivatives,
@2U
@2U
¼
:
@y @x @x @y
Thus
@M
@2U
¼
@y
@y @x
and
@N
@2U
¼
:
@x @x @y
@ U
Integrating both sides of @N
@x ¼ @x @y with respect to x,
we get
@U
N¼
þ f ðyÞ:
@y
Thus,
@U
@U
dx þ
þ f ð yÞ dy
Mdx þ Ndy ¼
@x
@y
@U
@U
¼
dx þ
dy þ f ð yÞdy
@x
@y
¼ d U þ f ð yÞdy
Z
¼ d½U þ f ð yÞdy
2
Thus,
M dx þ N dy is the exact differential of U þ
R
f ðyÞ dy and, hence, the differential equation M dx
þ N dy ¼ 0 is exact.
or
Z
Z
M dx þ
ðterms of N not containing xÞdy ¼ C:
y constant
EXAMPLE 1.31
Solve
ð2x cos y þ 3x2 yÞdx þ ðx3 x2 sin y yÞdy ¼ 0
Solution. Comparing with M dx þ N dy ¼ 0, we get
M ¼ 2x cos y þ 3x2 y;
N ¼ x3 x2 sin y y:
Then
@M
@N
¼ 2x sin y þ 3x2 ¼
:
@y
@x
Hence the given equation is exact. Therefore, the
solution of the equation is given by
Z
Z
M dx þ ðterms of N not containing xÞdy ¼ C
y constant
or
Z
Z
ð2x cos y þ 3x yÞdx þ
2
y constant
ydy ¼ C
Ordinary Differential Equations
or
or
y2
x2 cos y þ x3 y ¼ C:
2
or
Solution. Comparing with Mdx þ Ndy ¼ 0, we
observe that
or
M ¼ 2xy þ y tan y; and
Therefore,
@M
¼ 2x sec2 y þ 1;
@y
Z
x dx þ
dx ¼ 0
Z
y
cos t dtx2 y þ x ¼ C;
x2 ¼ t
y sin x2 x2 y þ x ¼ C:
@N
¼ 2x tan2 y:
@x
Solution. The given differential equation is
@N
@x
Comparing with M dx þ N dy ¼ 0, we get
ðy cos x þ sin y þ yÞdx þ ðsin x þ x cos y þ xÞdy ¼ 0
Hence,
¼
and so the equation is exact. Its
solution is given by
Z
Z
ð2xy þ y tan yÞ dx þ sec2 y dy ¼ C
y constant
M ¼ y cos x þ sin y þ y;
N ¼ sin x þ x cos y þ x;
and so
or
@M
¼ cos x þ cos y þ 1;
@y
@N
¼ cos x þ cos y þ 1:
@x
x2 y þ xy x tan y þ tan y ¼ C:
EXAMPLE 1.33
Solve
ð2xy cos x 2xy þ 1Þ dx þ ðsin x x Þ dy ¼ 0:
2
2
2
Solution. Comparing with M dx þ N dy ¼ 0, we note
that
M ¼ 2xy cos x2 2xy þ 1; N ¼ sin x2 x2 :
Then
@M
@N
¼ 2x cos x2 2x;
¼ 2x cos x2 2x:
@y
@y
@N
Thus @M
@y ¼ @x and, therefore, the given equation is
exact. The solution of the given equation is
Z
Z
ð2xy cos x2 2xy þ 1Þ dx þ 0 dy ¼ C
y constant
1.17
EXAMPLE 1.34
Solve
dy y cos x þ sin y þ y
þ
¼ 0:
dx sin x þ x cos y þ x
N ¼ x2 x tan2 y þ sec2 y:
@M
@y
2x cos x dx 2y
2
y
EXAMPLE 1.32
Solve
ð2xy þ y tan yÞdx þ ðx2 x tan2 y þ sec2 yÞdy ¼ 0:
Z
Z
n
Therefore, the given equation is exact and its solution is
Z
Z
ðy cos x þ sin y þ yÞ dx þ 0 dy ¼ C
y constant
or
y sin x þ x sin y þ xy ¼ C:
EXAMPLE 1.35
Solve
ðsec x tan x tan y ex Þdx þ sec x sec2 y dy ¼ 0:
Solution. Comparing with M dx þ N dy ¼ 0, we have
M ¼ sec x tan x tan y ex ;
N ¼ sec x sec2 y:
1.18
n
Engineering Mathematics-II
Therefore,
@M
¼ sec x tan x sec2 y;
@y
@N
¼ sec x tan x sec2 y:
@x
Hence the equation is exact and its solution is given
by
Z
Z
x
ðsec x tan x tan y e Þ dx þ 0 dx ¼ C
Solution. Comparing the given differential equation
with Mdx þ Ndy ¼ 0, we get
M ¼ ax þ hy þ g; N ¼ hx þ by þ f ;
@M
@N
¼ h;
¼ h:
@y
@x
Hence the given differential equation is exact and its
solution is
Z
Z
ðax þ hy þ gÞdx þ ðby þ f Þdy ¼ C
y constant
or
y constant
Z
or
sec x tan x dx ex ¼ C
tan y
or
tan y sec x ex ¼ 0
EXAMPLE 1.36
Solve
a
x2
y2
þ hyx þ gx þ b þ fy ¼ C
2
2
or
ax2 þ bx2 þ 2hxy þ 2gx þ 2fy þ k ¼ 0;
which represents a family of conics.
x
ð1 þ ex=y Þ dx þ ex=y ð1 Þ dy ¼ 0:
y
Solution. Comparing with M dx þ N dy ¼ 0, we get
x
x=y
x=y
N¼e
1 ;
M¼1þe ;
y
@M
x
x
¼ ex=y 2 ¼ 2 ex=y ;
@y
y
y
@N
x
1
x
x=y 1
x=y
¼e
1
þe
¼ 2 ex=y :
@x
y
y
y
y
Hence the given differential equation is exact and its
solution is
Z
Z
ð1 þ ex=y Þdx þ 0 dx ¼ C
y constant
1.12
EQUATIONS REDUCIBLE TO EXACT
EQUATION
Differential equations which are not exact can
sometimes be made exact on multiplying by a suitable factor called an integrating factor.
The integrating factor for Mdx þ Ndy ¼ 0 can
be found by the following rules:
1. If Mdx þ Ndy ¼ 0 is a homogeneous
equation in x and y, then
1
MxþNy
is an inte-
grating factor, provided Mx þ Ny 6¼ 0.
2. If the equation Mdx þ Ndy ¼ 0 is of the
1
form f (xy)y dx þ (xy)x dy ¼ 0, then MxNy
is an integrating factor, provided Mx – Ny 6¼
0.
3. Let M dx þ N dy ¼ 0 be a differential
or
xþy e
x=y
¼ C:
EXAMPLE 1.37
Show that the differential equation
ðax þ hy þ gÞdx þ ðhx þ by þ f Þdt ¼ 0
is the differential equation of a family of conics.
equation. If
ð@M=@yÞð@N=@xÞ
N
is
R a function
of x only, say f (x), then e f ðxÞ dx is an
integrating factor.
4. Let M dx þ N dy ¼ 0. If ð@N =@xÞð@M=@yÞ
M
R is a
function of y only, say f ( y), then e
is an integrating factor.
f ðyÞ dy
Ordinary Differential Equations
5. For the equation
0
0
xa yb ðmydxþnxdyÞþxa yb ðm0 ydxþn0 xdyÞ ¼ 0
the integrating factor is xh yk, where h and k
are such that
aþhþ1 bþk þ1
¼
;
m
n
a0 þhþ1 b0 þk þ1
¼
:
m0
n0
EXAMPLE 1.38
Solve
y dx x dy þ log x dx ¼ 0:
Solution. The given equation is not exact. Dividing
by x2, we get
y
x dy 1
dx 2 þ 2 log x dx ¼ 0
2
x
x
x
or
ydx xdy 1
þ 2 log x dx ¼ 0
x2
x
or
Z
y
1
log x dx ¼ 0
d
þd
x
x2
or
Z
d
1
y
log xdx 2
x
x
¼0
Thus x12 is an integrating factor and on integration,
we get the solution as
Z
1
y
log x dx ¼ C
x2
x
On integration by parts, we have
Z
log x
1
y
þ
dx ¼ C þ :
x
x2
x
or
1
1
y
log x ¼ C þ :
x
x
x
EXAMPLE 1.39
Solve
a2 ðx dy y dxÞ
xdx þ ydy ¼
:
x 2 þ y2
Solution. We know that
y
x dy y dx
:
d tan1
¼
x
x2 þ y2
n
1.19
Therefore, the given differential equation is
y
xdx þ ydy a2 d tan1
x
Integrating, we get
x2 y2
y
þ a2 tan1 ¼ C
2
2
x
or
y
x2 þ y2 2a2 tan1 ¼ k ðconstantÞ:
x
EXAMPLE 1.40
Solve
x dy y dx þ aðx2 þ y2 Þ dx ¼ 0:
Solution. Dividing throughout by x2 þ y2, we get
x dy
y dx
2
þ a dx ¼ 0
2
2
x þy
x þ y2
or
x dy y dx
þ adx ¼ 0
x2 þ y2
or
y
d tan1 þ adx ¼ 0:
x
Integrating, we get
y
tan1 þ ax ¼ C:
x
EXAMPLE 1.41
Solve
ðx2 y 2xy2 Þdx ¼ ðx3 3x2 yÞdy ¼ 0:
Solution. The given equation is homogeneous.
Comparing it with M dx þ N dy ¼ 0, we have
M ¼ x2 y 2xy2 ; N ¼ x3 þ 3x2 y;
@M
@N
¼ x2 4xy;
¼ 3x2 þ 6xy:
@y
@x
Therefore, @M
@y 6¼
not exact. Further
@N
@x and
so the given equation is
Mx þ Ny ¼ x3 y 2x2 y2 x3 y þ 3x2 y2 ¼ x2 y2 6¼ 0:
1
¼ x21y2 .
Hence the integrating factor is MxþNy
1
Multiplying the given equation by x2 y2 , we have
1 2
x
3
dx dy ¼ 0
ð15Þ
y x
y2 y
1.20
n
Since
@
@y
Engineering Mathematics-II
1 2
y x
¼
@
@x
x 3
y2 y
¼
1
;
y2
the equation (15) is exact and so its solution is
Z Z 1 2
3
dx dy ¼ C
y x
y
y constant
or
x
2 log x þ 3 log y ¼ C:
y
EXAMPLE 1.42
Solve
y dx þ 2x dy ¼ 0:
Solution. The given equation is of type Mdx þNdy ¼ 0
and is homogeneous. Further
Solution. The given equation is homogeneous and
comparing with Mdx þ N dy ¼ 0, we get
M ¼ x2 y; N ¼ x3 y3 :
Then
Mx þ Ny ¼ x3 y x3 y y4 ¼ y4 6¼ 0:
Thus the integrating factor is y1 4 . Multiplying the
given differential equation throughout by – y14 , we get
1
1
4 x2 ydx þ 4 ðx3 þ y3 Þdy ¼ 0
y
y
or
3
x2
x
1
þ dy ¼ 0;
3 dx þ
y
y4 y
which is exact. Hence the required solution is
Z
Z
1
1
2
dy ¼ C
3 x dx þ
y
y
M ¼ y; N ¼ 2x;
@M
@N
@M
@N
¼ 1;
¼ 2;
6¼
:
@y
@x
@y
@x
Thus the equation is not exact. But
Mx þ N y ¼ xy þ 2xy ¼ 3xy 6¼ 0:
EXAMPLE 1.44
Solve
1
is the integrating factor. Multiplying
Therefore, 3xy
1
, we get
the given equation throughout by 3xy
1
2x
y dx þ
dy ¼ 0
3xy
3xy
or
1
2
dx þ dy ¼ 0:
3x
3y
The solution is
Z
Z
1 1
2 1
dx þ
dy ¼ C
3 x
3 y
or
1
2
log x þ log y ¼ C
3
3
or
log xy2 ¼ k ¼ log p
or
yðxy þ 2x2 y2 Þ dx þ xðxy x2 y2 Þ dy ¼ 0:
Solution. The given differential equation is of the
form
f ðxyÞydx þ ðxyÞxdy ¼ 0:
Also comparing with M dx þ Ndy ¼ 0, we get
M ¼ yðxy þ 2x2 y2 Þ; N ¼ xðxy x2 y2 Þ:
Therefore,
Mx Ny ¼ 3x3 y3 6¼ 0:
Thus, 3x13 y3 is the integrating factor. Multiplying
throughout by 3x13 y3 , we have
1
1
ðyðxyþ2x2 y2 ÞÞdxþ 3 3 ½xðxyx2 y2 Þdy ¼ 0:
3
3
3x y
3x y
or
xy ¼ p ðconstantÞ:
2
EXAMPLE 1.43
Solve
Since
x2 y dx ðx3 þ y3 Þ dy ¼ 0:
x3
þ log y ¼ C:
3y3
1
2
þ
3x2 y 3x
1
1
dx þ
x dy ¼ 0:
3xy2 3y
@
1
2
@
1
1
þ
¼
;
@y 3x2 y 3x
@x 3xy2 3y
Ordinary Differential Equations
the above equation is exact and its solution is
Z Z 1
2
1
þ
dx
þ
dy ¼ C
3x2 y 3x
3y
n
1.21
EXAMPLE 1.46
Solve
ðx2 y2 þ xy þ 1Þ y dx þ ðx2 y2 xy þ 1Þ x dy ¼ 0:
y constant
or
1
2
1
þ log x log y ¼ C
3xy 3
3
or
1
þ 2 log x log y ¼ k ðconstantÞ:
xy
Solution. The given differential equation is of the form
f ðxyÞydx þ ðxyÞxdx ¼ 0:
Moreover, comparing the given equation with
M dx þ N dy ¼ 0, we get
M ¼ x2 y3 þ xy2 þ y; N ¼ x3 y2 x2 y þ x:
Therefore
Mx Ny ¼ x3 y3 þ x2 y2 þ xy x3 y3 þ x2 y2 xy
EXAMPLE 1.45
Solve
ð1 þ xyÞ y dx þ ð1 xyÞ x dy ¼ 0:
Solution. The given differential equation is of the
form
f ðxyÞ ydx þ ðxyÞ x dy ¼ 0:
Comparing with M dx þ Ndy ¼ 0, we have
M ¼ y þ xy2 ; N ¼ x x2 y:
Therefore,
¼ 2x2 y2 6¼ 0:
Therefore, the integrating factor is 2x12 y2 . Multiplying
the given differential equation throughout by 2x12 y2 ,
we get
1
ðx2 y3 þ xy2 þ yÞ dx
2x2 y2
1
þ
ðx3 y2 x2 y þ xÞ dy ¼ 0
2x2 y2
or
y 1
1
1
1
1
þ þ 2 dx þ
x þ 2 dy ¼ 0;
2 2x 2x y
2
y xy
Mx Ny ¼ 2x2 y2 6¼ 0:
Therefore, the integrating factor is 2x12 y2 . Multiplying
the given differential equation throughout by 2x12 y2 ,
we get
1
1
ðy þ xy2 Þdx þ 2 2 ðx x2 yÞ ¼ 0
2
2
2x y
2x y
which is exact. Hence the solution of the equation is
Z Z y 1
1
1
1
dx þ
þ þ
dy ¼ C
2 2x 2x2 y
2
y
or
or
1
1
þ
2
2x y 2x
1
1
dx þ
2
2xy
2y
dy ¼ 0:
We note that this equation is exact. Hence its solution is
Z Z
1
1
1
þ
dx þ
dy ¼ C
2x2 y 2x
2y
y constant
or
y constant
1
1
1
1
þ log x log y ¼ C
2y
x
2
2
or
x 1
¼ k ðconstantÞ:
log y xy
xy 1
1
1
þ log x log y ¼ k
2 2
2yx
2
or
xy 1
x
þ log ¼ k ðconstantÞ:
xy
y
EXAMPLE 1.47
Solve
yð2xy þ 1Þ dx þ xð1 þ 2xy x3 y3 Þdy ¼ 0:
Solution. The differential equation in question is of
the form
f ðxyÞydx þ ðxyÞxdy ¼ 0:
1.22
n
Engineering Mathematics-II
Further comparing the given equation with M dx þ
N dy ¼ 0, we get
M ¼ 2xy2 þ y; N ¼ x þ 2x2 y x4 y3 :
R
Therefore, e 1 dx ¼ ex is the integrating factor.
Multiplying the given differential equation
throughout by ex, we get
ðx2 þ y2 þ 2xÞex dx þ 2yex dy ¼ 0;
Therefore,
Mx Ny ¼ 2x2 y2 þ xy xy 2x2 y2 x4 y4
¼ x4 y4 :
Thus the integrating factor is x14 y4 . Multiplying
the given differential equation throughout by x41y4 ,
we get
1
1
ð2xy2 þ yÞdx 4 4 ðx þ 2x2 y x4 y3 Þ dy ¼ 0:
x4 y4
x y
or
2
1
1
2
1
3 2 4 3 dx þ 3 4 2 3 þ dy ¼ 0;
x y
x y
x y
x y
y
which is exact. Hence the required solution is
Z
Z
ðx2 þ y2 þ 2xÞ ex dx þ
0 dy ¼ C;
y constant
which yield
ðx2 þ y2 Þ ex ¼ C:
EXAMPLE 1.49
Solve
1=3
ðxy2 ex Þdx x2 y dy ¼ 0:
Solution. Comparing the given equation with M dx þ
N dy ¼ 0, we get
1=3
which is exact. Hence the solution of the equation is
Z
Z 2
1
1
dx
þ
dy ¼ C
3
2
4
3
x y
x y
y
y constant
M ¼ xy2 ex ; N ¼ x2 y;
@M
@N
¼ 2xy;
¼ 2xy:
@y
@x
Therefore,
@M
@y
or
1
1
þ 3 3 þ log y ¼ C:
2
2
x y
3x y
@N
@x
N
2xy ð2xyÞ
4
¼ ;
x2 y
x R
¼
EXAMPLE 1.48
Solve
ðx2 þ y2 þ 2xÞ dx þ 2y dy ¼ 0:
which is a function of x only. Hence e x dx ¼
e4 log x ¼ x14 is the integrating factor. Multiplying
the equation throughout by x14 , we get
2
y
1 x1=3
y
dx 2 dy ¼ 0;
e
3
4
x
x
x
Solution. Comparing the given equation with Mdx þ
Ndy ¼ 0, we get
which is exact. The required solution is, therefore,
Z 2
y
1 x1=3
dx ¼ C
e
x3 x4
y constant
M ¼ x2 þ y2 þ 2x; N ¼ 2y
or
which gives
@M
@N
¼ 2y;
¼ 0:
@y
@x
@N
@x
N
2y
¼
¼ 1 ¼ x0 ðfunction of xÞ:
2y
Z
or
y2
1
2þ
2x
3
Thus the equation is not exact. We have
@M
@y
y2
1
þ
2x2 3
3 x1=3
e dx ¼ C
x4
Z
or
1
et dt ¼ C; t ¼ x3 :
y2
1
þ et ¼ C
2x2 3
4
Ordinary Differential Equations
or
2
3y
1=3
þ 2ex ¼ k ðconstantÞ:
2
2x
Solution. Comparing with M dx þ N dy ¼ 0, we get
M ¼ xy3 þ y;
@M
¼ 3xy2 þ 1;
@y
@M
@y
N ¼ 2x2 y2 þ 2x þ 2y4 ;
@N
¼ 4xy2 þ 2:
@x
@N
@x
@M
@y
@N @M
@x @y
¼
xy2 þ1
yðxy2 þ1Þ
¼
ðfunctionof y aloneÞ:
M
R1
Hence, the integrating factor is e y dy ¼ elog y ¼ y.
Multiplying the given equation throughout by y, we
get
1
y
yðxy3 þ yÞdx þ 2yðx2 y2 þ x þ y4 Þdy ¼ 0
2
2 3
or
y constant
2
4x
y þ 2 dx þ x þ 2y 3 dy ¼ 0;
y
y
which is exact. The solution of the given differential
equation is, therefore,
Z
Z 2
y þ 2 dx þ 2y dy ¼ 0
y
y constant
or
2
y þ 2 x þ y2 ¼ C:
y
5
which is exact. Therefore, its solution is
Z
Z
ðxy4 þ y2 Þdx þ y5 dy ¼ C
3y3 6 3ðy2 þ 2Þ
3
¼
¼ ;
4
3
y þ 2y
yðy þ 2Þ
y
1 4
1
ðy þ 2yÞdx þ 3 ðxy3 þ 2y4 4xÞdy ¼ 0
3
y
y
ðxy þ y Þdx þ ð2x y þ 2xy þ y Þdy ¼ 0;
4
¼
which is a function ofR y alone. Therefore, the inte3
grating factor is e y dy ¼ e3 log y ¼ y3 ¼ y13 .
Multiplying the given differential equation
throughout by y13 , we get
@N
@x ,
Since
6¼
the equation is not exact. It is also
not homogeneous. It is also not of the form f (x y) y
dx þ (x y)x dy ¼ 0. We note that
1.23
which shows that the given equation is not exact.
Further,
M
EXAMPLE 1.50
Solve
ðxy3 þ yÞdx þ 2ðx2 y2 þ x þ y4 Þ dy ¼ 0:
n
EXAMPLE 1.52
Solve
yðxy þ 2x2 y2 Þdx þ xðxy x2 y2 Þdy ¼ 0:
or
x2 y4
y6
þ xy2 þ ¼ C:
2
6
EXAMPLE 1.51
Solve
ðy4 þ 2yÞdx þ ðxy3 þ 2y4 4xÞdy ¼ 0:
Solution. The given differential equation is neither
homogeneous nor is of the type f (x y) y dx þ (x y)
x dy ¼ 0. Comparing with Mdx þ Ndy ¼ 0, we get
M ¼ y4 þ 2y; N ¼ xy3 þ 2y4 4x:
Therefore,
@M
¼ 4y3 þ 2;
@y
Solution. The given equation can be written as
xyð ydx þ xdyÞ þ x2 y2 ð2ydx xdyÞ ¼ 0:
Comparing it with
0
0
xa yb ðmydx þ nxdyÞ þ xa yb ðm0 y dx þ n0 x dyÞ ¼ 0;
we note that a ¼ b ¼ 1, a0 ¼ b0 ¼ 2, m ¼ n ¼ 1 and
m0 ¼ 2, n0 ¼ –1. Then the integrating factor is xhyk,
where
aþhþ1 bþkþ1
¼
;
m
n
a0 þ h þ 1 b0 þ k þ 1
¼
;
m0
n0
that is,
@N
¼ y3 4;
@x
hþ2 kþ2
¼
;
1
1
3þh 3þk
¼
2
1
1.24
n
Engineering Mathematics-II
or
h þ 2 ¼ k þ 2;
3 k ¼ 6 þ 2k
or
h k ¼ 0;
h þ 2k ¼ 9:
Solving for h and k, we get h ¼ k ¼ –3. Hence, the
integrating factor is x31y3 . Multiplying throughout by
1
x3 y3 , we get
1
2
1
1
þ
dx
þ
dy ¼ 0;
x2 y x
xy2 y
which is exact. Therefore, the solution is
Z Z 1
2
1
þ
dx þ
dy ¼ C
2
x y x
y
y constant
or
Solving for h and k, we get h ¼ 52 k ¼ 12.
Hence, x5=21y1=2 is the integrating factor. Multiplying
the given differential equation by x5=21y1=2 , we get
ðx5=2 y3=2 þ 2x1=2 y1=2 Þdx
þ ð2x1=2 y1=2 þ 2x3=2 y1=2 Þdy ¼ 0;
which is exact. Therefore, the required solution is
Z
Z
5=2 3=2
1=2 1=2
ðx
y þ 2x
y Þdx þ 0 dy ¼ 0
y constant
or
2
y3=2 x3=2 þ 4y1=2 x1=2 ¼ C
3
or
y3=2 x3=2 þ 6x1=2 y1=2 ¼ C
1
1
þ 2 log x log y ¼ C
y
x
or
pffiffiffiffiffi y3=2
6 xy ¼ C:
x
or
1
þ 2 log x log y ¼ C:
xy
EXAMPLE 1.53
Solve
EXAMPLE 1.54
Solve
ð3xy 2ay2 Þ þ ðx2 2axyÞ dy ¼ 0:
Solution. The given equation can be written as
ðy þ 2x yÞ dx þ ð2x xyÞ dy ¼ 0:
2
2
3
xy0 ð3ydx þ xdyÞ þ yx0 ð2aydx 2axdyÞ ¼ 0:
Comparing this with
Solution. The given equation can be written as
x yðy dx x dyÞ þ x y ð2y dx þ 2x dyÞ ¼ 0:
0
2 0
Comparing this with
0
0
we get
a ¼ 1; b ¼ 0;
0
xa yb ðmydx þ nxdyÞ þ xa yb ðm0 ydx þ n0 xdyÞ ¼ 0;
we get
a ¼ 0; b ¼ 1; a0 ¼ 2; b0 ¼ 0;
m ¼ 1; n ¼ 1; m0 ¼ 2; n0 ¼ 2:
Then the integrating factor is xhyk, where
a þ h þ 1 b þ k þ 1 a0 þ h þ 1 b0 þ k þ 1
¼
;
¼
;
m
n
m0
n0
or
hþ1 2þk 3þh kþ2
¼
;
¼
1
1
2
2
or
h þ k ¼ 3 and 2h 2k ¼ 4:
0
xa yb ðmydx þ nxdyÞ þ xa yb ðm0 ydx þ n0 xdyÞ ¼ 0;
a0 ¼ 0; b0 ¼ 1;
m ¼ 3; n ¼ 1; m0 ¼ 2a; n0 ¼ 2a:
Then the integrating factor is xh yk, where
a þ h þ 1 b þ k þ 1 a0 þ h þ 1 b0 þ k þ 1
¼
;
¼
m
m
m0
n0
or
2þh kþ1 hþ1 kþ2
¼
;
¼
3
1
2a
2a
or
h 3k ¼ 1; h k ¼ 1:
Thus h ¼ 1, k ¼ 0. Therefore, the integrating factor
is x. Multiplying the given differential equation
Ordinary Differential Equations
throughout by x, we get
ð3x2 y 2axy2 Þdx þ ðx3 2ax2 yÞdy ¼ 0:
@M
@y
ð2x3=7 y5=7 þ x11=7 y12=7 Þdx
ðx10=7 y12=7 3x4=7 y10=7 Þdy ¼ 0:
This transformed
equation is exact and its solution is
Z
ð2x3=7 y5=7 þ x11=7 y12=7 Þ dx ¼ C
@N
@x
¼
and so the transformed equation is
Hence
exact.
The required solution is
Z
Z
ð3x2 y 2axy2 Þdx þ 0 dy ¼ 0
y constant
y constant
or
7 10=7 5=7 7 4=7 12=7
x
y
¼ C:
x y
5
4
1.13
or
yx3 ax2 y2 ¼ C:
EXAMPLE 1.55
Solve
ð2x2 y2 þ yÞdx þ ðx3 y þ 3xÞ dy ¼ 0:
Solution. Comparing the given differential equation
with Mdx þ Ndy ¼ 0, we get
M ¼ 2x2 y2 þ y; N ¼ x3 y þ 3x;
@M
¼ 4x2 y þ 1;
@y
APPLICATIONS OF FIRST ORDER AND FIRST
DEGREE EQUATIONS
The aim of this section is to form differential equations for physical problems like flow of current in
electric circuits, Newton law of cooling, heat flow
and orthogonal trajectories, and to find their solutions.
(A) Problems Related to Electric Circuits
Consider the RCL circuit shown in the Figure 1.3
and consisting of resistance, capacitor, and inductor
connected to a battery.
@N
¼ 3 3x2 y:
@x
@N
Since @M
@y 6¼ @x , the given equation is not exact.
However, the given equation can be written in the
form
x2 yð2ydx xdyÞ þ x0 y0 ðydx þ 3xdyÞ ¼ 0:
1.25
y–19/7. Multiplying the given equation throughout
by x–11/7 y–19/7, we get
Let
2
2
3
2
Then M ¼ 3x y 2axy ; N ¼ x 2ax y:
@M
@N
¼ 3x2 4axy;
¼ 3x2 4axy:
@y
@x
n
L
E
R
C
Comparing it with
0
0
xa yb ðmy dx þ nx dyÞ þ xa yb ðm0 ydx þ n0 xdyÞ ¼ 0;
we get
a ¼ 2; b ¼ 1; a0 ¼ 0; b0 ¼ 0;
m ¼ 2; n ¼ 1; m0 ¼ 1; n0 ¼ 3:
The integrating factor is xh yk, where
aþhþ1 bþkþ1
a0 þ h þ 1 b0 þ k þ 1
¼
¼
;
m
n
m0
n0
or
h þ 2k ¼ 7; 3h k ¼ 2:
Solving these equations for h and k, we get h ¼ 11
7
–11/7
and k ¼ 19
7 . Thus the integrating factor is x
Figure 1.3
We know that the resistance is measured in ohms,
capacitance in farads, and inductance in henrys. Let
I denote the current flowing through the circuit and
Q denote the charge. Since the current is rate of flow
V
of charge, we have I ¼ dQ
dt . Also, by Ohm’s law, I ¼
R (resistance). Therefore, the voltage drop across a
resistor R is RI. The voltage drop across the inductor
Q
L is L dI
dt and the voltage drop across a capacitor is C .
If E is the voltage (e.m.f.) of the battery, then by
Kirchhoff’s law, we have
dI
Q
ð16Þ
L þ R I þ ¼ EðtÞ;
dt
C
1.26
n
Engineering Mathematics-II
where L, C, and R are constants. Since I ¼ dQ
dt , we
Rt
have Q ¼ IðuÞ du and so (16) reduces to
0
Zt
dI
1
L
þ RI þ
IðuÞ du ¼ EðtÞ:
dt
C
0
The forcing function (input function), E(t), is supplied by the battery (voltage source). The system
described by the above differential equation is
known as harmonic oscillator.
The equation (16) can be written as
dI R
Q
E
þ Iþ
¼ ðtÞ
dt L
LC
L
EXAMPLE 1.57
Find the time t when the current reaches half of its
theoretical maximum in the circuit of Example 1.56.
Solution. From Example 1.56, we have
E
ð1 eRt=L Þ:
R
The maximum current is ER. By the requirement of
the problem, we must have
I¼
1E
E
¼ ð1 eRt=L Þ
2R
R
ð17Þ
or
EXAMPLE 1.56
Given that I ¼ 0 at t ¼ 0, find an expression for the
current in the LR circuit shown in the Figure 1.4.
or
1
¼ eRt=L
2
Rt
1
¼ log ¼ log 2
L
2
L
or
t ¼
L
log 2:
R
E
R
EXAMPLE 1.58
In an LR circuit, an e.m.f. of 10 sin t volts is applied.
If I(0) ¼ 0, find the current in the circuit.
Figure 1.4
Solution. By Kirchhoff’s law, we have
dI R
E
þ I¼ ;
dt L
L
ð18Þ
which is Leibnitz’s
linear equation. Its integrating
RR
of (18) is
factor is e ZL dt ¼ eRt=L . Hence the solution
Z
E
E
eRt=L dt þ C ¼
eRt=L dt þ C
I eRt=L ¼
L
L
¼
E eRt=L
E Rt=L
þC ¼
e
þ C:
L R=L
R
Thus
E
ð19Þ
þ CeRt=L :
R
But I(0) ¼ 0, therefore, (19) yields 0 ¼ ER þ C and
so C ¼ ER. Hence
I ¼
E E Rt=L E
¼ ð1 eRt=L Þ:
e
R R
R
Clearly I increases with time t and attains its maximum value, E/R.
I ¼
Solution. In an LR circuit the current is governed by
the differential equation
dI R
E
þ I¼ :
dt L
L
We are given that E ¼ 10 sin t. Therefore,
dI R
10
þ I ¼
sin t:
dt L
L
This is Leibnitz’s
linear equation with integrating
RR
factor as e L dt ¼ eRt=L : Therefore, its solution is
Z
10
Rt=L
Ie
¼
sin t eRt=L dt þ C
ZL
ð20Þ
10
eRt=L sin t dt þ C
¼
L
But we know (using integration by parts) that
Z
eat
ða sin bt b cos btÞ
eat sin bt dt ¼ 2
a þ b2
eat
b
¼ 2
sin bt tan1 :
2
a þb
a
Ordinary Differential Equations
Therefore (20) reduces to
I eRt=L
10
e
R
¼
sin
t
cos
t
þC
L ðR2 =L2 Þ þ 1 L
Rt=L
10L2 eRt=L
¼ 2
ðR sin t L cos tÞ þ C
L
R2 þ L2
10 eRt=L
¼ 2
ðR sin t L cos tÞ þ C:
R þ L2
Hence
10
ðR sin t L cos tÞ þ CeRt=L :
I ¼ 2
R þ L2
Using the initial condition I(0) ¼ 0, we get
C ¼ R210L
þL2 . Hence
10
ðR sin t L cos t þ LeRt=L Þ:
I¼ 2
R þ L2
EXAMPLE 1.59
If voltage of a battery in an LR circuit is E0 sin t,
find the current I in the circuit under the initial
condition I(0) ¼ 0.
Solution. Proceeding as in Example 1.58, we get
n
1.27
Using Kirchhoff’s law, we have
Zt
1
RI þ
IðuÞ du ¼ E0 sin vt:
C
0
Differentiating both sides with respect to t, we have
dI I
R
þ ¼ vE0 sin vt
dt C
or
dI
I
vE0
cos vt;
þ
¼
R
dt RC
which is Leibnitz’s
linear equation. The integrating
R 1
t
dt
RC
RC
factor is e
¼ e . Therefore, the solution of the
above first order equation is
Z
vE0
t
t
RC
cos vt:eRC dt þ C
Ie ¼
R
Z
v E0
t
cos vt eRC dt þ C
¼
R
R
ax
Using eax cos bx dx ¼ a2eþb2 ða cos bx þ b sin bxÞ,
we have
"
#
t
vE0
eRC
1
Ie ¼
cosvt þvsinvt þC
1 2
R
þv2 RC
t
RC
I ¼
E0
E0 L
E0 R
eRt=L þ 2
cos t þ 2
sin t:
R 2 þ L2
R þ L2
R þ L2
RC
t
vR2 C 2 E0
eRC
¼
ðcosvt þRCvsinvtÞ þC
2
R C
1þR2 v2 C 2
t
EXAMPLE 1.60
Find the current in the following electric circuit containing condenser C, resistance R, and a battery of
e.m.f. E0 sin vt with the initial condition I(0) ¼ 0.
Solution. The RC circuit of the problem is shown in
Figure 1.5.
E 0 sin t
R
¼
vCE0 eRC
ðcosvt þRCvsinvtÞþC:
1þR2 v2 C 2
vCE0
But I(0) ¼ 0 implies C ¼ 1þR
2 v2 C 2 :
Hence
vCE0
I¼
ðcos vt þ RCv sin vt et=RC Þ:
1 þ R2 v 2 C 2
EXAMPLE 1.61
A voltage E e–at is applied at t ¼ 0 to a LR circuit.
Find the current at any time t.
Solution. The differential equation governing the LR
circuit is
C
Figure 1.5
dI R
e:m:f : E eat
:
þ I¼
¼
L
dt
L
L
1.28
n
Engineering Mathematics-II
Rt
As in Example 1.58, the integrating factor is e L .
Therefore, the solution of the above equation is
Z
Z
E eat Rt
E
Rt
Rt
L
L
e dt þ C ¼
I e ¼
e L at dt þ C
L
L
E e L at
E
Rt
½e L at þ C
þC ¼
R
L La
R aL
Using the initial condition Q(0) ¼ 0, we get
C ¼ 45. Therefore,
4
8
4
Q ¼ cos 2t þ sin 2t et :
5
5
5
Rt
¼
Hence
I¼
and so
I¼
E
Rt
eat þ C e L :
R aL
Using the initial condition I(0) ¼ 0, we get
R
. Hence
C ¼ RaL
I¼
E
Rt
½eat e L :
R aL
EXAMPLE 1.62
An RC circuit has an e.m.f. given in volt by 400 cos
2t, a resistance of 100 ohms, and a capacitance of
10–2 Farad. Initially there is no charge on the
capacitor. Find the current in the circuit at any time t.
(B) Problems Related to Newton’s Law of Cooling
Newton’s law of cooling states that the time rate of
change of the temperature of a body is proportional
to the temperature difference between the body and
its surrounding medium.
Let T be the temperature of the body at any
time t and T0 be the temperature of the surrounding
at that particular time. Then, according to Newton’s
Law of Cooling,
dT
1 ðT T0 Þ
dt
and so
dT
¼ kðT T 0Þ;
dt
Solution. The equation governing the circuit is
Q
¼ E:
RI þ
C
We are given that R ¼ 100 ohms, C ¼ 10–2 farad
and E ¼ 400 cos 2t. Thus, we have
I þ Q ¼ 4 cos 2t
or
dQ
dQ
þ Q ¼ 4 cos 2t;
R since I ¼ dt :
dt
The integrating factor is e 1 dt ¼ et . Therefore, the
solution is
Z
Q:et ¼ 4 cos 2t et dt
et
½cos 2t þ 2 sin 2t þ C
5
4 t
8
e cos 2t þ et sin 2t þ C:
¼
5
5
¼ 4
Thus
dQ
8
16
4
¼ sin 2t þ cos 2t þ et :
dt
5
5
5
the negative sign with the constant of proportionality is required to make dT
dt negative in cooling process
when T is greater than T0, and positive in a heating
process when T is less than T0.
Equation (21) is first order differential equation
and can be solved for T.
EXAMPLE 1.63
A body at a temperature of 50 F is placed outdoors,
where the temperature is 100 F. If after 5 minutes,
the temperature of the body is 60 F, find the time
required by the body to reach a temperature
of 75 F.
Solution. By Newton’s law of cooling, we have
dT
¼ kðT T0 Þ
dt
or
Q¼
4
8
cos 2t þ sin 2t þ C et :
5
5
ð21Þ
dT
þ k T ¼ k T0 :
dt
n
Ordinary Differential Equations
But when t ¼ 0, T ¼ 80 C. Therefore, C ¼ 80 – 40
¼ 40 and we have
But T0 ¼ 100 F. Therefore,
dT
þ kT ¼ 100k;
dt
R
which is linear. The integrating factor is e k dt ¼ ekt .
Hence the solution is
Z
Z
kt
kt
kt
T:e ¼ 100 k e dt þ C ¼ 100e þ C
Now when t ¼ 20, T ¼ 60 . Therefore
or
or
T¼ Ce
kt
þ 100:
When t ¼ 0, T ¼ 50, therefore, C ¼ –50. Hence
T 40 ¼ 40ekt :
20 ¼ 40 e20 k or e20 k ¼
T 40 ¼ 40eð20
1
Now it is given that T ¼ 60 at t ¼ 5. Hence
log 2Þ t
:
When t ¼ 40, we have
4
:
5
T 40 ¼ 40e2 log 2
or
Taking log, we get
1
1
4
1
k ¼ log ¼ ð0:223Þ ¼ 0:045:
5
5
5
Hence T ¼ –50e0.045t þ 100. When T ¼ 75, we get
e0:045 t ¼ 12, which yields 0:045t ¼ log 12 and so
t ¼ 15.4 minutes.
EXAMPLE 1.64
A body originally at 80 C cools down to 60 C in
20 minutes, the temperature of the air being 40 C.
Find the temperature of the body after 40 minutes
from the original.
Solution. By Newton’s law of cooling, we have
dT
¼ kðT T0 Þ
dt
and so variable separation gives
dT
¼ kdt:
T T0
Integrating, we have
logðT T0 Þ ¼ kt þ log C
or
T T0 ¼ C ekt
1
2
1
1
20 k ¼ log ; which yields k ¼ log 2:
2
20
Hence
T ¼ 50ekt þ 100:
60 ¼ 50e5k þ 100 or e5k ¼
1.29
or T 40 ¼ C ekt :
T ¼ 40 þ 40 elog4 ¼ 40 þ
40
¼ 50 C:
4
(C) Problems Relating to Heat Flow
The fundamental principles of heat conduction are:
(i) Heat always flow from a higher temperature to a lower temperature.
(ii) The quantity of heat in a body is proportional to its mass and temperature (Q ¼
mst), where m is the mass, s is the specific
heat, and t is the time.
(iii) The rate of heat flow across an area is proportional to the area and to the rate of
change of temperature with respect to its
distance normal to the area.
Let Q be the quantity of heat flow per second across
a slab of area A and thickness x and whose faces
are kept at temperature T and T þ T. Then, by the
above principles
dT
QµA :
dx
or
dT
Q ¼ k A
ð22Þ
dx
1.30
n
Engineering Mathematics-II
where k is a constant, called the coefficient of
thermal conductivity and depends upon the material
of the body. Negative sign has been taken since T
decreases as x increases. The relation (22) is called
the Fourier’s law of conductivity.
EXAMPLE 1.65
The inner and outer surfaces of a spherical shell are
maintained at temperature T0 and T1, respectively. If
the inner and outer radii of the shell are r0 and r1,
respectively and k is the thermal conductivity, find the
amount of heat lost from the shell per unit time. Also
find the temperature distribution through the shell.
Solution. We have
Q ¼ k ð4x2 Þ
dT
;
dx
where x is the radius. Thus
dT ¼ Q dx
: :
4k x2
Integrating, we get
Q
1
Q
þC ¼
þC
T¼
4k
x
4xk
ð23Þ
Q
Q
þ C and T1 ¼
þ C:
4r0 k
4r1 k
Subtracting, we get
Q 1
1
Q r1 r0
¼
T0 T1 ¼
4k r0 r1
4k r0 r1
or
4k r0 r1 ðT0 T1 Þ
:
r1 r0
When x ¼ r0, T ¼ T0, then (23) gives
Q
þC
T0 ¼
4r0 k
4kr0 r1 ðT0 T1 Þ
þC
¼
4 r0 kðr1 r0 Þ
r1 ðT0 T1 Þ
þ C:
¼
r1 r0
Q¼
r1 ðT0 T1 Þ
r1 r0
T0 ðr1 r0 Þ r1 ðT0 T1 Þ r1 T1 r0 T0
¼
¼
r1 r0
r1 r0
Hence (23) transforms to
C ¼ T0 4kr0 r1 ðT0 T1 Þ r1 T1 r0 T0
þ
r1 r0
4xkðr1 r0 Þ
1
ðT0 T1 Þr0 r1
þ r1 T1 r0 T0 :
¼
r1 r0
x
T¼
EXAMPLE 1.66
A spherical shell of inner and outer radii 10 cm and
15 cm, respectively, contains steam at 150 C. If the
temperature of the outer surface of the shell is 40 C
and thermal conductivity k ¼ 0.0025, find the
temperature half-way through the thickness of the
shell under steady state conditions.
Solution. With the notation of Example 1.65, we have
r0 ¼ 10cm; r1 ¼ 15cm; x ¼ 12:5cm;
Now T ¼ T0 when x ¼ r0 and T ¼ T1 when x ¼ r1.
Therefore, we have
T0 ¼
Thus
T0 ¼ 150 C; T1 ¼ 40 C:
Hence
1
ðT0 T1 Þr0 r1
þ r1 T1 r0 T0
x
r1 r0
1 ð150 40Þ150
þ 600 1500
¼
5
12:5
1 16500
1
900 ¼ ½1320 900 ¼ 84 C:
¼
5 12:5
5
T¼
EXAMPLE 1.67
A long hollow pipe has a inner radius of r0 cm and
outer radius of r1 cm. The inner surface is kept at a
temperature T0 and the outer surface at the temperature T1. If thermal conductivity is k, find the
heat lost per second of 1 cm length of the pipe. Also
find the temperature distribution through the
thickness of the pipe.
Solution. Let Q cal/sec be the constant quantity of
heat flowing out radially through the surface of the
pipe having radius x cm and length 1 cm. Then the
area of the lateral surface is 2x. Therefore, by
Ordinary Differential Equations
dT
dT
¼ kð2xÞ
dx
dx
Solution.Using the notation of Example 1.67, we
have
and so
dT ¼ Q dx
: :
2k x
r0 ¼ 10cm; r1 ¼ 15 cm; x ¼ 12:5;
T0 ¼ 150 C; T1 ¼ 40 C; k ¼ 0:0025:
Integrating, we get
T ¼
Q
log x þ C:
2k
When x ¼ r0, T ¼ T0, and so
Q
log r0 þ C
T0 ¼ 2k
When x ¼ r1, T ¼ T1 and we have
Q
log r1 þ C:
T1 ¼ 2k
ð24Þ
Hence using (28), we have
40 150
12:5
log
10
log 15
10
1:25
¼ 89:5 C:
¼ 150 110 log
1:5
T ¼ 150 þ
ð25Þ
ð26Þ
Subtracting (26) from (25), we have
Q
Q
r1
T0 T1 ¼
½log r0 log r1 ¼
log :
r0
2k
2k
Thus,
Q ¼ 2kðT0 T1 Þ
log rr10 ;
ð27Þ
which gives the heat lost per second in 1 cm of the
pipe. Further subtracting (25) from (24) and using
(27), we get
Q
Q
x
½log x log r0 ¼ log
2k
2k
r0
2kðT0 T1 Þ
x
¼
log
2k log rr10
r0
EXAMPLE 1.69
A long hollow pipe has an inner diameter of 10 cm
and outer diameter of 20 cm. The inner surface is
kept at 200 C and the outer surface at 50 C. The
thermal conductivity is 0.12. How much heat will
be lost per second from a portion of 1 cm of the pipe
and what is the temeprature at a distance of 7.5 cm
from the centre of the pipe?
Solution. From Example 1.67, we have
Hence,
ðT1 T0 Þ
x
log :
r1
log r0
r0
T ¼ T0 þ
ðT1 T0 Þ
x
log
log rr10
r0
2kðT0 T1 Þ
log rr10
2kð200 50Þ 300k
¼
¼
log 2
log 10
5
300ð0:12Þ
¼ 163cal=sec:
¼
log 2
Q¼
T T0 ¼ ¼
1.31
temperature half-way through the covering under
steady-state conditions.
Fourier law,
Q ¼ kA
n
Also
ðT1 T0 Þ
x
log
log rr10
r0
ð50 200Þ
7:5
¼ 200 þ
log
log 2
5
1:5
¼ 200 150ð:58Þ ¼ 113 C:
¼ 200 150 log
2
T ¼ T0 þ
ð28Þ
which gives the required temperature distribution
through the thickness of the pipe.
(D) Rate Problems
EXAMPLE 1.68
A pipe 20 cm in diameter contains steam at 150 C
and is protected with a covering 5 cm thick. If
thermal conductivity is 0.0025 and the temperature
of the outer surface of the covering is 40 C, find the
In some problems, the rate at which a quantity changes is
a known function of the amount present and/or the time
and it is desired to find the quantity itself. Radioactive
nuclei decay, population growth, and chemical reactions
are some of the phenomenon of this kind.
1.32
n
Engineering Mathematics-II
Let x be the amount of radioactive nuclei present after t years. Then dx
dt represents the rate of
decay. Since the nuclei decay at a rate proportional
to the amount present, we have
dx
¼ kx;
ð29Þ
dt
where k is constant of proportionality.
The law of chemical reaction states that the rate
of change of chemical reaction is proportional to
the amount of substance present at that instant.
Thus, the differential equation (29) governs the
chemical reaction of first order.
Moreover, if the rate at which amount of a substance increases or decreases is found to be jointly
proportional to two factors, each factor being a linear function of x, then the chemical reaction is said
to be of second order. For example, if a solution
contains two substances whose amounts at the
beginning are a and b respectively and if equal
amount x of each substance changes in time t, then
the amounts of the substance left in the solution at
time t are (a – x) and (b – x) and, therefore, we have
dx
¼ kða xÞðb xÞ:
dt
Taking the case of population growth, we assume
that the population is a continuous and differentiable function of time. Let x be the number of individuals in a population at time t. Then rate of
change of population is proporitional to the number
of individuals in it at any time. Thus equation (29) is
valid for population growth also.
EXAMPLE 1.70
If 10% of 50 mg of a radioactive material decays in
2 hours, find the mass of the material left at any
time t and the time at which the material has
decayed to one-half of its initial mass.
Solution. Let x denote the amount of material present
at time t. Then the equation governing the decay is
dx
¼ kx:
dt
Variable separation gives
dx
¼ kdt:
x
Integrating, we get
log x ¼ kt þ log C
or
x ¼ Cekt :
At t ¼ 0, x ¼ 50. Therefore, C ¼ 50 and so x ¼ 50
ekt. At t ¼ 2, 10% of the mass present is decayed.
Thus 5 mg of the substance has been decayed and
45 gm still remains. Therefore, 45 ¼ 50 e2k. Thus
1
45
k ¼ log ¼ 0:053:
2
50
Hence mass of the material left at any time t is
x ¼ 50 e0:053 t :
Further, when half of the material is decayed, we
have x ¼ 25 mg and so
25 ¼ 50 e0:053 t
or
0:053t ¼ log
1
2
or
t ¼ 13 hours:
EXAMPLE 1.71
A tank contains 1000 litres of fresh water. Salt water
which contains 150 gm of salt per litre runs into it at
the rate of 5 litres/min and well stirred mixture runs
out of it at the same rate. When will the tank contain
5000 gm of salt?
Solution. Let x denote the amount of salt in the tank at
time t. Then
dx
¼ IN OUT:
dt
The brine flows at the rate of 5 litres/min and each
litre contains 150 gm of salt. Thus
IN ¼ 5 150 ¼ 750 gm=min
Since the rate of outflow equals the rate of inflow, the
tank contains 1000 litres of mixture at any time t.
This 1000 litres contains x gm of salt at time t and so
x
gm/litres.
the concentration of the salt at time t is 1000
Since mixture flows out at the rate of 5 litres/min,
we have
x
x
5 ¼
gm=litres:
OUT ¼
1000
200
Ordinary Differential Equations
Thus the differential equation for x becomes
dx
x
¼ 750 :
ð30Þ
dt
200
Since initially there was no salt in the tank, we have
the initial condition x(0) ¼ 0. The equation (30) is
linear and separable. We have in fact
dx
dt
¼
:
150000 x
200
Integrating, we get
t
þ C:
ð31Þ
logð150000 xÞ ¼
200
Using the initial condition x(0) ¼ 0, we have
C ¼ log 150000:
Hence (31) yields
t
¼ log 150000 logð150000 xÞ
200
or
150000
t ¼ 200 log
:
150000 x
If x ¼ 5000 gm, then
150000
30
t ¼ 200 log
¼ 200 log ¼ 6:77min:
145000
29
EXAMPLE 1.72
If the population of a city gets doubled in 2 years and
after 3 years the population is 15,000, find the initial
population of the city.
Solution. Let x denote the population at any time t
and let x0 be the initial population of the city. Then
dx
¼ kx;
dt
which has the solution as
x ¼ C ekt :
At t ¼ 0, x ¼ x0. Hence C ¼ x0. Thus
x ¼ x0 ekt
But at t ¼ 2, x ¼ 2x0. Therefore
2x0 ¼ x0 e2k or e2k ¼ 2
ð32Þ
1
log 2 ¼ 0:347
2
1.33
Hence (32) reduces to
¼ x0 e0:347 t :
x
At t ¼ 3, x ¼ 15,000 and so
15; 000 ¼ x0 eð0:347Þ ð3Þ ¼ x0 ð2:832Þ
Hence
x0 ¼
15000
¼ 5297:
2:832
(E) Falling Body Problems
Consider a body of mass m falling under the influence of gravity g and an air resistance, which is
proportional to the velocity of the falling body.
Newton’s second law of motion states that the net
force acting on a body is equal to the time rate of
change of the momentum of the body.
Thus
d
F ¼ ðmvÞ:
dt
If m is assumed to be constant, then
dv
F ¼ m ;
ð33Þ
dt
where F is the net force on the body and v is the
velocity of the body at time t.
The falling body is under the action of two
forces: (i) Force due to gravity which is given by
the weight mg of the body (ii) the force due to
resistance of air and that is –kv, where k 0 is a
constant of proportionality. Thus (33) yields
dv
mg kv ¼ m
dt
or
dv
k
þ v ¼ g;
ð34Þ
dt
m
which is equation of motion for the falling body.
If air resistance is negligible, then k ¼ 0 and we
have
dv
¼ g:
ð35Þ
dt
The differential equation (35) is separable and we
have
dv ¼ g dt:
or
k¼
n
Integrating, we get
v ¼ gt þ C:
1.34
n
Engineering Mathematics-II
But when t ¼ 0, v ¼ 0 and so C ¼ 0. Hence
v ¼ gt:
ð36Þ
Also velocity is time rate of change of displacement
x and so
dx
¼ gt or dx ¼ gt dt:
dt
Integrating, we get
1
x ¼ gt2 þ k ðconstantÞ:
2
But at t ¼ 0, the displacement is 0. Therefore k ¼ 0.
Hence
1
ð37Þ
x ¼ gt2 :
2
EXAMPLE 1.73
A body of mass 16 kg is dropped from a height of 625
ft. Assuming that there is no air resistance, find the
time required by the body to reach the ground.
Solution. By (37), we have (with g ¼ 32 ft/sec2)
1
x ¼ gt2 ¼ 16t2 ;
2
therefore,
t
2
rffiffiffiffiffiffiffiffi
rffiffiffiffiffi
x
625
25
¼
¼
¼
¼ 6:25 sec:
16
16
4
(F) Orthogonal Trajectories
Recall that a curve which cuts every member of a
given family of curves according to some definite
law is called a trajectory of the family.
A curve which cuts every member of a given
family of curves at right angles is called orthogonal
trajectories. Further, two families of curves are said
to be orthogonal if every member of either family
cuts each member of the other family at right angles.
Consider a one-parameter family of curves in
the xy-plane defined by
f ðx; y; cÞ ¼ 0;
ð38Þ
where c denotes the parameter. Differentiating (38),
with respect to x and eliminating c between (38) and
the resulting equation, we get the differential
equation of the family in question. Let the differential equation be
dy
¼ 0:
ð39Þ
F x; y;
dx
To obtain the equation of the orthogonal trajectory,
dx
equawe replace dy
dx by dy and get the differential
tion of orthogonal trajectory as F x; y; dx
dy :
Solution of this differential equation will yield the
equation of the orthogonal trajectory.
In case of polar curves
f ðr; h; cÞ ¼ 0:
ð40Þ
Differentiating (40) and eliminating c between (40)
and the resulting equations, we get the differential
equation of the family represented by (40). Let the
differential equation be
dr
ð41Þ
F ðr ; h ; Þ ¼ 0:
dh
dr
by r2 dh
Replacing dh
dr in (41), we get the differential equation of the orthogonal trajectory as
dh
Þ ¼ 0:
ð42Þ
dr
Solution of (42) will then yield the equation of the
required orthogonal trajectory.
F ðr ; h r2
EXAMPLE 1.74
Find the orthogonal trajectories of the family of
curves x2 þ y2 ¼ cx .
Solution. We have
ð43Þ
x2 þ y2 cx ¼ 0:
Differentiating, we get
dy
¼ c:
ð44Þ
2x þ 2y
dx
Eliminating c between (43) and (44) yields
dy
x2 þ y2
2x þ 2y
¼
x
dx
or
dy
y2 x2
:
ð45Þ
¼
2xy
dx
The equation (45) is the differential equation of the
family represented by (43). Therefore, the differential equation of the orthogonal trajectory is
dy
2xy
:
ð46Þ
¼ 2
dx
x y2
This is an homogeneous equation. Substituting
y ¼ vx, and separating variables, we get
dx
1
2v
dv ¼ 0:
þ þ 2
x
v
v þ1
Ordinary Differential Equations
log x log v þ logðv þ 1Þ ¼ C
2
or
xðv2 þ 1Þ ¼ kv; ðC ¼ log kÞ:
Substituting v ¼ yx ; we get
x2 þ y2 ¼ ky:
EXAMPLE 1.75
Find the orthogonal trajectory of the family of the
curves xy ¼ C.
Solution. The equation of the given family of curves is
xy ¼ C:
ð47Þ
Differentiating, we get
dy
þy
dx
1.35
EXAMPLE 1.76
Find the orthogonal trajectories of the family of
curves y ¼ ax2.
Integrating, we get
x
n
¼ 0
Solution. The given family represented by the equation
y ¼ ax2
ð50Þ
is a family of parabolas symmetric about y-axis with
vertices at (0, 0). Differentiating with respect to x,
we get
dy
¼ 2ax:
ð51Þ
dx
Eliminating a between (50) and (51), we get
dy
2xy
2y
¼ 2 ¼
:
dx
x
x
Therefore, differential equation of the orthogonal
trajectory is
dy
x
¼ ð52Þ
dx
2y
or
2y dy þ x dx ¼ 0:
or
dy
y
¼ :
ð48Þ
dx
x
Therefore, the differential equation of the family of
orthogonal trajectory is
dy
x
¼
ð49Þ
dx
y
or
x dx y dy ¼ 0:
Integrating, we get
2
y2 x2
þ
¼ C
2
2
or
x2 y2
þ ¼ C:
ð53Þ
2
1
The orthogonal trajectories represented by (53) are
ellipses (shown in the Figure 1.7)
y
Integrating, we get
x2 y2 ¼ k;
which is the equation of orthogonal trajectories
called equipotential lines (shown in Figure 1.6).
y
0
x
Figure 1.7
x
0
Figure 1.6
EXAMPLE 1.77
Show that the system of confocal and coaxial parabolas is self-orthogonal.
1.36
n
Engineering Mathematics-II
Solution. The equation of the family of confocal
parabolas having x-axis as their axis is of the form
y2 ¼ 4aðx þ aÞ:
Differentiating, we get
dy
y
¼ 2a
dx
or
dy
2a
¼ :
dx
y
ð54Þ
ð55Þ
From (55), we have a ¼ 2y dy
dx : Substituting this
value in (54), we get
dy
1 dy
2
þ xþ y
y ¼ 2y
dx
2 dx
or
2
dy
dy
þ2x y ¼ 0;
y
dx
dx
ð56Þ
which is the differential equation of the given family.
dx
Replacing dy
dx by dy in (56) we obtain (56) again.
Hence, each member of family (54) cuts every other
member of the same family orthogonally.
EXAMPLE 1.78
Find the orthogonal
trajectories of the family of
2
2
curves ax2 þ b2yþl ¼ 1.
Solution. The equation of the family of the given
curve is
x2
y2
¼ 1:
ð57Þ
þ
a2 b2 þ l
Differentiating with respect to x, we get
x
y dy
þ 2
¼ 0:
2
a
b þ l dx
ð58Þ
From (57) and (58) we have respectively
a2 y 2
a2 y dy
2
b2 þ l ¼ 2
:
and
b
þ
l
¼
a x2
x dx
Hence
a2 y 2
a2 y dy
¼
a2 x 2
x dx
or
dy
xy
:
¼
dx a2 x2
Therefore, differential equation of the orthogonal
trajectory is
dy a2 x2
¼
xy
dx
or
a2 x2
a2
y dy ¼
dx ¼ dx x dx:
x
x
Integrating, we get
y2
x2
¼ a2 log x þ C
2
2
or
x2 þ y2 ¼ 2a2 log x þ kðconstantÞ;
which is the equation of the required orthogonal
trajectories.
EXAMPLE 1.79
Find the orthogonal trajectory of the cardioid
r ¼ a(1 – cos h).
Solution. The equation of the family of given cardioide is
r ¼ að1 cos hÞ:
ð59Þ
Differentiating with respect to h, we have
dr
¼ a sin h:
ð60Þ
dh
Dividing (60) by (59), we get the differential
equation of the given family as
1 dr
sin h
2 sin h=2 cos h=2
h
¼ cot ð61Þ
¼
¼
r dh 1 cos h
2
2 sin2 h=2
dr
Replacing dh
by r2 dh
in (61), we get
dr
1
dh
h
r2
¼ cot
r
dr
2
or
dr
h
þ tan dh ¼ 0;
ð62Þ
r
2
which is the equation of the family of orthogonal
trajectories. Integrating (62), we get
log r 2 log cos h ¼ log C
2
or
h
h
log r ¼ log C þ log cos2 ¼ log C cos2
2
2
Ordinary Differential Equations
or
h
¼ C ð1 þ cos hÞ;
r ¼ C cos
2
which is the equation of orthogonal trajectory of the
given family.
2
EXAMPLE 1.80
Find the orthogonal trajectories of the family of
curves x2 þ y2 ¼ c2
Solution. The equation of the given family of curves is
x2 þ y2 c2 ¼ 0
ð63Þ
Differentiating (63), we get
dy
¼0
2x þ 2y
dx
or
dy
x
¼ ;
ð64Þ
dx
y
which is the differential equation representing the
given curves. Therefore the differential equation
of the required family of orthogonal trajectories is
dy y
¼
dx x
or
dy dx
¼
ð65Þ
y
x
Integrating, we get
log y ¼ log x þ log k
or
y ¼ kx;
which is the equation of the orthogonal trajectories,
which are straight lines through the origin as shown
in the Figure 1.8.
y
x
0
Figure 1.8
EXAMPLE 1.81
Find the orthogonal trajectories of the curves r2 ¼
a2 cos 2h.
n
1.37
Solution. We are given that
r2 ¼ a2 cos 2h ¼ a2 ð1 2 sin2 hÞ:
Differentiating (66) w.r.t. h, we have
dr
2r
¼ 2a2 sin 2h:
dh
Dividing (67) by (66), we get
2 dr 2 sin 2h
¼
:
r dh
cos 2h
dr
with –r2 dh
Replacing dh
dr , we get
dh
sin 2h
¼
¼ 2 tan 2h
2r
dr
cos 2h
or
dr
¼ cot 2h:
r
Integrating, we get
1
log r ¼ log sin 2h þ log C
2
or
2 log r ¼ 2 log C þ log sin 2h
or
log r2 ¼ log C 2 þ log sin 2h
or
r2 ¼ C 2 sin 2h:
1.14
ð66Þ
ð67Þ
LINEAR DIFFERENTIAL EQUATIONS
Definition 1.17. A differential equation in which the
dependent variable and its derivatives occur only in
the first degree and are not multiplied together is
called a linear differential equation.
Thus, a linear differential equation of nth order
is of the form
dny
d n1 y
d n2 y
a0 n þa1 n1 þa2 n2 þ...þan y ¼ FðxÞ ð68Þ
dx
dx
dx
where a0, a1, . . ., an and F(x) are functions of x alone.
If a0, a1, . . ., an are constants, then the above
equation is called a linear differential equation with
constant coefficients.
If F is identically zero, then the equation (68)
reduces to
dny
d n1 y
d n2 y
a0 n þ a1 n1 þ a2 n2 þ ::: þ an y ¼ 0 ð69Þ
dx
dx
dx
and is called a homogeneous linear differential
equation of order n.
1.38
n
Engineering Mathematics-II
Definition 1.18. If f1, f2, . . ., fn are n given functions
and c1, c2, . . ., cn are n constants, then the
expression
c1 f1 þ c2 f2 þ . . . þ cn fn
Then
Dn u þ a1 Dn1 u þ a2 Dn2 u þ ::: þ an u
¼ Dn ðc1 y1 þ c2 y2 þ ::: þ cn yn Þ
þ a1 Dn1 ðc1 y1 þ c2 y2 þ ::: þ cn yn Þ
is called a linear combination of f1, f2 , . . ., fn .
þ a2 Dn2 ðc1 y1 þ c2 y2 þ ::: þ cn yn Þ
þ . . . þ an ðc1 y1 þ c2 y2 þ ::: þ cn yn Þ
Definition 1.19. The set of functions { f1, f2, . . ., fn} is
said to be linearly independent on [a, b] if the relation
c1 f1 þ c2 f2 þ . . . þ cn fn ¼ 0
¼ c1 ðDn y1 þ a1 Dn1 y1 þ ::: þ an y1 Þ
for all x 2 [a, b] implies that
c1 ¼ c2 ¼ . . . ¼ cn ¼ 0:
þ . . . þ cn ðDn yn þ a1 Dn1 yn þ ::: þ an yn Þ
¼ 0 þ 0 þ :: þ 0 ¼ 0 using ð71Þ:
d
Definition 1.20. The symbol D ¼ dx
is called a difd2
d3
2
3
ferential operator. Similarly, D ¼ dx
2 , D ¼ dx3 , . . .,
dn
Dn ¼ dx
n are also regarded as operators. In terms of
these symbols, the equation (68) takes the form
ða0 Dn þ a1 Dn1 þ ::: þ an1 D þ an Þy ¼ FðxÞ
or
f ðDÞy ¼ FðxÞ;
where
f ðDÞ ¼ a0 Dn þ a1 Dn1 þ . . . þ an1 D þ an :
Dðl uÞ ¼ lDðuÞ
m n
ðD D Þ ðuÞ ¼ Dmþn ðuÞ
ðDm Dn Þ ðuÞ ¼ ðDn Dm Þ ðuÞ:
Theorem 1.4. Any linear combination of linearly
independent solutions of the homogeneous linear
differential equation is also a solution (in fact,
complete solution) of that equation.
Proof: Let y1, y2, . . ., yn be the solution of the
homogeneous linear differential equation
ð70Þ
9
Dn y1 þa1 Dn1 y1 þa2 Dn2 y1 þ:::þan y1 ¼0 >
>
=
Dn y2 þa1 Dn1 y2 þa2 Dn2 y2 þ:::þan y2 ¼0
ð71Þ
:::::::::::::
>
>
;
Dn yn þa1 Dn1 yn þa2 Dn2 yn þ:::þan yn ¼0
Let
u ¼ c1 y1 þ c2 y2 þ . . . þ cn yn :
EXAMPLE 1.82
Show that c1 sin x þ c2 cos x is a solution of
y ¼ 0.
d2 y
dx2
þ
y1 ¼ sin x;
y2 ¼ cos x:
dy1
¼ cos x;
dx
dy2
¼ sin x
dx
d 2 y2
¼ cos x:
dx2
Then
Dðu þ vÞ ¼ Du þ Dv
Therefore,
Hence, u ¼ c1 y1 þ c2 y2 þ . . . þ cn yn is also a solution
of the homogeneous linear differential equation (68).
Since this solution contains n arbitrary constants, it is
a general or a complete solution of (70).
Solution. Let
We note that
ðDn þ a1 Dn1 þ a2 Dn2 þ ::: þ an Þy ¼ 0
þ c2 ðDn y2 þ a1 Dn1 y2 þ ::: þ an y2 Þ
d 2 y1
¼ sin x;
dx2
We note that
d 2 y1
þ y1 ¼ sin x þ sin x ¼ 0
dx2
and
d 2 y2
þ y2 ¼ cos x þ cos x ¼ 0:
dx2
Hence, sin x and cos x are solutions of the given
equation. These two solutions are linearly independent. Therefore, their linear combination c1 sin x
þ c2 cos x is also a solution of the given equation.
Theorem 1.5. If y1 is a complete solution of the
homogeneous equation f (D)y ¼ 0 and y2 is a particular solution containing no arbitrary constants of the
differential equation f (D)y ¼ F(x), then y1 þ y2 is
the complete solution of the equation f (D)y ¼ F(x).
Ordinary Differential Equations
Proof: Since y1 is a complete solution of the homogeneous differential equation f (D)y ¼ 0, we have
ð72Þ
f ðDÞy1 ¼ 0
Further, since y2 is a particular solution of linear
differential equation f (D)y ¼ F(x), we have
ð73Þ
f ðDÞy2 ¼ FðxÞ:
Adding (70) and (71), we get
f ðDÞy1 þ f ðDÞy2 ¼ FðxÞ
f ðDÞðy1 þ y2 Þ ¼ FðxÞ:
Hence, y1 þ y2 satisfies the equation f (D)y ¼ F(x)
and so is the complete solution since it contains n
arbitrary constants.
Definition 1.21. Let f (D)y ¼ F(x) be a linear differential equation with constant coefficients. If y1 is a
complete solution of f (D)y ¼ 0 and y2 is a particular
solution of f (D)y ¼ F(x), then y1 þ y2 is a complete
solution of f (D)y ¼ F(x) and then y1 is called the
complementary function and y2 is called the particular
integral of the differential equation f (D)y ¼ F(x) .
Consider the homogeneous differential equation
f (D)y ¼ 0. Then
Let y ¼ e
mx
The symbolic form of this equation is
ðDn þ a1 Dn1 þ ::: þ an1 D þ an Þy ¼ 0;
ð76Þ
where a1, a2, . . ., an are constants. If y ¼ e is a
solution of (76), then
ð77Þ
mn þ a1 mn1 þ ::: þ an1 m þ an ¼ 0:
mx
Three cases arise, according as the roots of (77)
are real and distinct, real and repeated or complex.
ð74Þ
Suppose that the auxiliary equation (77) has n
distinct roots m1, m2, . . ., mn. Therefore, (77)
reduces to
ð78Þ
ðm m1 Þðm m2 Þ . . . ðm mn Þ ¼ 0
Equation (78) will be satisfied by the solutions of
the equations
ðD m1 Þy ¼ 0; ðD m2 Þy ¼ 0; . . . ;
ðD mn Þ ¼ 0:
We consider (D – m1)y ¼ 0. This can be written as
dy
m1 y ¼ 0;
dx
which is linear differential equation with integrating
factor as em1 x . Therefore, its solution is
Z
0 : em1 x dx þ c1
y:em1 x ¼
be a solution of (74). Then
Dy ¼ me ; D2 y ¼ m2 emx ; . . . ;
mx
Dn y ¼ mn emx
and so (74) transforms to
ðmn þ a1 mn1 þ ::: þ an Þem x ¼ 0:
Since emx 6¼ 0, we have
mn þ a1 mn1 þ a2 mn2 þ ::: þ an ¼ 0:
ð75Þ
mx
It follows, therefore, that if e is a solution of f (D)y
¼ 0, then equation (75) is satisfied.
The equation (75) is called auxiliary equation
for the differential equation f (D)y ¼ 0.
1.15
1.39
Case I. Distinct Real Roots
or
þ ::: þ an Þy ¼ 0:
ðDn þ a1 Dn1 þ an2
2
n
SOLUTION OF HOMOGENEOUS LINEAR
DIFFERENTIAL EQUATION WITH CONSTANT
COEFFICIENTS
Consider the homogeneous linear differential equation
dny
d n1 y
dy
þ
a
þ ::: þ an1 þ an y ¼ 0:
1
dxn
dxn1
dx
or
y ¼ c1 em1 x :
Similarly,
the solution of ðD m2 Þy ¼ 0 is c2 em2 x ;
the solution of ðD m3 Þy ¼ 0 is c3 em3 x ;
...
...
the solution of ðD my Þy ¼ 0 is cn emn x :
Hence, the complete solution of homogeneous differential equation (76) is
ð79Þ
y ¼ c1 em1 x þ c2 em2 x þ . . . þ cn emn x :
Case II. Repeated Real Roots
Suppose that the roots m1 and m2 of the auxiliary
equation are equal. Then the solution (79) becomes
y ¼ c1 em1 x þ c2 em1 x þ c3 em3 x þ . . . þ cn emn x
¼ ðc1 þ c2 Þ em1 x þ c3 em3 x þ . . . þ cn emn x :
1.40
n
Engineering Mathematics-II
solution will come out to be
Since it contains n – 1 arbitrary constants, it is not a
complete solution of the given differential equation.
We shall show that the part of the solution corresponding to equal roots m1 and m2 is (c1x þ c2) em1 x .
To prove it, consider the equation
In general, if m1 ¼ m2 ¼ . . . ¼ mk, then the complete
solution of the differential equation shall be
ðD m1 Þ2 y ¼ 0;
y ¼ ðc1 xk1 þ c2 xk2 þ ::: þ ck Þ þ ckþ1 emkþ1 x þ :::
y ¼ ðc1 x2 þ c2 x þ c3 Þem1 x þ c4 em4 x þ . . . þ cn emn x :
þcn emn x :
that is,
ðD m1 ÞðD m1 Þy ¼ 0:
Substituting (D – m1)y ¼ U, the above equation
becomes
ðD m1 ÞU ¼ 0
or
dU
m1 U ¼ 0
dx
or
dU
¼ m1 dx:
U
Integrating, we get
log U ¼ m1 x þ log C1
or
log
and so
U
U
¼ m1 x or ¼ em1 x
c1
c1
U ¼ c1 e
m1 x
:
Hence
ðD m1 Þy ¼ c1 em1 x
Case III. Conjugate Complex Roots
(a)
Suppose that the auxiliary equation has a nonrepeated complex root a þ ib. Then, since the
coefficients are real, the conjugate complex
number a – ib is also a non-repeated root.
Thus, the solution given in (79) becomes
y¼c1 eðaþibÞx þc2 eðaibÞx þc3 em3 x þ:::þcn emn x
¼eax ðc1 eibx þc2 eibx Þþc3 em3 x þ:::þcn emn x
¼eax ½c1 ðcosbxþisinbxÞþc2 ðcosbxisinbxÞ
þc3 em3 x þ:::þcn emn x
¼eax ½ðc1 þc2 Þcosbxþiðc1 c2 Þsinbx
þc3 em3 x þ:::þcn emn x
¼eax ½k1 cosbxþk2 sinbxþc3 em3 x þ:::þcn emn x ;
where k1 ¼ c1 þ c2, k2 ¼ i(c1 – c2) .
(b) If two pairs of imaginary roots are equal, then
m1 ¼ m2 ¼ a þ ib
and
m3 ¼ m4 ¼ a ib:
Using Case II, the complete solution is
or
dy
m1 y ¼ c1 em1 x ;
dx
which is again a linear equation with integrating
factor em1 x . Hence, the solution is
Z
m1 x
y: e
¼
c1 em1 x :em1 x þ c2
¼ c1 x þ c2 ;
which yields
¼ eax ½ðc1 x þ c2 Þ cos bx þ ðc3 x þ c4 Þ sin bx
þc5 em5 x þ . . . þ cn emn x :
EXAMPLE 1.83
Solve
d3y d2y
dy
þ 2 þ 4 þ 4y ¼ 0:
3
dx
dx
dx
Solution. The symbolic form of the given equation is
y ¼ ðc1 x þ c2 Þem1 x :
Hence, the complete solution of the given differential equation is
y ¼ ðc1 x þ c2 Þem1 x þ c3 em3 x þ . . . þ cn emn x :
Remark 1.1. If three roots of the auxiliary equation
are equal, that is, m1 ¼ m2 ¼ m3, then the complete
ðD3 þ D2 þ 4D þ 4Þy ¼ 0:
Therefore its auxiliary equation is
m3 þ m2 þ 4m þ 4 ¼ 0:
By inspection –1 is a root. Therefore, (m þ 1) is a
factor of m3 þm2 þ 4m þ 4. The synthetic division
Ordinary Differential Equations
by m þ 1 gives
1.41
Solution. The auxiliary equation is
1
1
1
4
1
0
m4 5m3 þ 6m2 þ 4m 8 ¼ 0;
4
1 0
4
whose roots are 2, 2, 2, and –1. Hence the general
solution is
0
y ¼ ðc1 þ c2 x þ c3 x2 Þe2x þ c4 ex :
4
Therefore, the auxiliary equation is
EXAMPLE 1.86
2
Solve ddxy2 2 dy
dx þ 10y ¼ 0 subject to the conditions
ðm þ 1Þðm2 þ 4Þ ¼ 0
and so
m ¼ 1 and m ¼ 2i:
Hence, the complementary solution is
y ¼ c1 ex þ e0 x ðc2 cos 2x þ c3 sin 2xÞ
y(0) ¼ 4, y0 (0) ¼ 1.
Solution. The symbolic form of the given differential
equation is
¼ c1 ex þ c2 cos 2x þ c3 sin 2x:
EXAMPLE 1.84
Solve
d3y
d2y
dy
3
þ 3 y ¼ 0:
dx3
dx2
dx
Solution. The symbolic form of the given equation is
ðD2 2D þ 10Þy ¼ 0:
Therefore, the auxiliary equation is
m2 2m þ 10 ¼ 0;
which yields
m¼1
y ¼ ex ðc1 cos 3x þ c2 sin 3xÞ:
Now,
Therefore, the auxiliary equation is
y0 ¼ c1 ½ex cos 3x 3ex sin 3x
m3 3m2 þ 3m 1 ¼ 0:
þ c2 ½ex sin 3x þ 3ex cos 3x
By inspection 1 is a root. Then synthetic division yields
1
1
3i:
Therefore, the solution is
ðD3 3D2 þ 3D 1Þy ¼ 0:
1
n
¼ ex cos 3xðc1 þ 3c2 Þ þ ex sin xðc2 3c1 Þ:
3
3
1
The initial conditions y(0) ¼ 4 and y0 (0) ¼ 1 yield
1
2
1
4 ¼ c1 and 1 ¼ c1 þ 3c2
2
1
0
and so c1 ¼ 4 and c2 ¼ –1. Hence the solution is
y ¼ ex ð4 cos 3x sin 3xÞ:
Therefore, the auxiliary equation is
ðm 1Þðm2 2m þ 1Þ ¼ 0 or ðm 1Þ3 ¼ 0:
Hence the roots are 1, 1, 1 and so the solution of the
given equation is
EXAMPLE 1.87
Solve
d3y
þ y ¼ 0:
dx3
y ¼ ðc1 þ c2 x þ c3 x Þe :
2
x
EXAMPLE 1.85
Find the general solution of
d4y
d3y
d2y
dy
5 3 þ 6 2 þ 4 8y ¼ 0:
4
dx
dx
dx
dx
Solution. The auxiliary equation is
m3 þ 1 ¼ 0
or
ðm þ 1Þðm2 m þ 1Þ ¼ 0:
1.42
n
Engineering Mathematics-II
pffiffi
Thus the roots are –1, 1 2 3 i. Hence, the general
solution of the equation is pffiffiffi
pffiffiffi
3
3
1
x
x
2
y ¼ c1 e þ e ðc2 cos
x þ c3 sin
xÞ:
2
2
EXAMPLE 1.88
Solve
d4y
¼ m4 y:
dx4
Solution. The auxiliary equation for the given differential equation is
s4 m4 ¼ 0
or
ðs þ mÞðs mÞðs2 þ m2 Þ ¼ 0
constant coefficients so that complete solution of
the equation may be found.
Definition 1.22. D1 F(x) is that function of x which when
operated upon by D yields F(x).
1
F(x) is that function of x, free
Similarly, f ðDÞ
from arbitrary constant, which when operated upon
by f (D) yields F(x).
1
is
Thus, D1 is the inverse operator of D and f ðDÞ
the inverse operator of f (D).
Theorem 1.6. D1 F(x) ¼
R
FðxÞ dx.
Proof: Let
1
and so s ¼ m, –m, ± mi. Hence the solution is
FðxÞ ¼ v:
D
y ¼ c1 emx þ c2 emx þ c3 cos mx þ c4 sin mx
¼ c1 ½cosh mx þ sinh mx þ c2 ½cosh mx sinh mx Operating both sides of (80) by D, we get
1
þ c3 cos mx þ c4 sin mx
D: FðxÞ ¼ Dv
D
¼ ðc1 þ c2 Þ cosh mx þ ðc1 c2 Þ sinh mx
þ c3 cos mx þ c4 sin mx
or
¼ C1 cosh mx þ C2 sinh mx þ c3 cos mx
dv
FðxÞ ¼ Dv ¼
þ c4 sin mx;
dx
where
C1 ¼ c1 þ c2 and C2 ¼ c1 c2 :
ð80Þ
or
dv ¼ FðxÞdx:
EXAMPLE 1.89
3
2
Solve 4 ddx3y þ 4 ddx2y þ dy
dx ¼ 0.
Integrating, we get
Solution. The auxiliary equation for the given differential equation is
4m3 þ 4m2 þ m ¼ 0
or
mð4m2 þ 4m þ 1Þ ¼ 0:
where no constant of integration is added since
1
DF(x) contains no constant. Thus,
Z
1
FðxÞ ¼
FðxÞ dx:
D
Thus the roots are m ¼ 0, –12, and –12. Hence the
solution is
y ¼ c1 e0 x þ ðc2 þ c3 xÞ ex=2
Theorem 1.7.
¼ c1 þ ðc2 þ c3 xÞex=2 :
1.16
COMPLETE SOLUTION OF LINEAR
DIFFERENTIAL EQUATION WITH CONSTANT
COEFFICIENTS
We now discuss the methods of finding particular
integral of a linear differential equation with
Z
v¼
FðxÞ dx;
Hence, D1 stands for integration.
1
Da FðxÞ
¼ eax
R
FðxÞeax dx.
Proof: Let
1
FðxÞ ¼ y:
ð81Þ
Da
Operating both sides of (81) by D – a, we have
FðxÞ ¼ ðD aÞy ¼ Dy ay
dy
¼ ay:
dx
n
1.43
1 ax
1 ax
e ¼
e ; provided f ðaÞ 6¼ 0:
f ðDÞ
f ðaÞ
ð83Þ
Ordinary Differential Equations
1
yields
Operating both sides by f ðDÞ
Therefore,
dy
ay ¼ FðxÞ;
dx
which is a linear differential equation with integrating factor e–ax. Therefore, its solution is
Z
ax
FðxÞ eax dx
y:e
or
y ¼ eax
Z
FðxÞ eax dx
or
1
FðxÞ ¼ eax
Da
FðxÞ e
ax
dx:
Proof: The given linear differential equation is
f ðDÞy ¼ FðxÞ
ð82Þ
1
f ðDÞF(x)
Hence
If f (a) ¼ 0, then D – a is a factor of f (D). So, let
Z
1
Theorem 1.8. f ðDÞ
F(x) is the particular integral of
f (D)y ¼ F(x).
Substituting y ¼
1
½ f ðaÞeax f ðDÞ
1 ax
¼ f ðaÞ
e :
f ðDÞ
eax ¼
in (82), we have F(x) ¼
1
F(x), which is true. Hence y ¼ f ðDÞ
F(x) is a solution
of (82).
1.16.1 Standard Cases of Particular Integrals
Consider the linear differential equation
f ðDÞ ¼ ðD aÞðDÞ:
ð84Þ
Then
1 ax
1
1
1 ax
e ¼
eax ¼
e
f ðDÞ
ðDaÞðDÞ
ðDaÞ ðDÞ
1
1 ax
e usingð83ÞsinceðaÞ 6¼ 0
¼
Da ðaÞ
1
1 ax
e
¼
ðaÞ Da
Z
1 ax
¼
eax :eax dx; by Theorem 1:7
e
ðaÞ
eax
ð85Þ
¼x
ðaÞ
f ðDÞy ¼ FðxÞ:
By Theorem 1.8, its particular integral is
1
P:I ¼
FðxÞ:
f ðDÞ
Differentiating (84) with respect to D gives
f 0 ðDÞ ¼ ðDÞ þ ðD aÞ0 ðDÞ:
Putting D ¼ a, we get f0 (a) ¼ (a). Therefore, (85)
reduces to
Case I. When F(x) ¼ eax
We have
f ðDÞ ¼ Dn þa1 Dn1 þa2 Dn2 þ...þan :
Therefore,
1 ax
eax
¼x
e ¼x
f ðDÞ
ðaÞ
¼x
f ðDÞeax ¼ ðDn þa1 Dn1 þ...þan Þeax
eax
d
dD f ðDÞ D¼a
eax
; provided f 0 ðaÞ 6¼ 0:
f 0 ðaÞ
ð86Þ
¼ Dn eax þa1 Dn1 eax þ...þan1 Deax þan eax If f 0 (a) ¼ 0, then the rule can be repeated to give
¼ an eax þa1 an1 eax þ...þan1 aeax þan eax
¼ ða þa1 a
¼ f ðaÞeax :
n
n1
þ...þan1 aþan Þe
ax
1 ax
eax ax
e ;
e ¼ x2 00
f ðaÞ
f ðDÞ
provided f 00 ðaÞ 6¼ 0 and so on:
1.44
n
Engineering Mathematics-II
Case II. When F(x) ¼ sin (ax þ b) or cos (ax þ b)
We have
x:
D sinðax þ bÞ ¼ a cosðax þ bÞ
D2 sinðax þ bÞ ¼ a2 sinðax þ bÞ
D3 sinðax þ bÞ ¼ a3 cosðax þ bÞ
D4 sinðax þ bÞ ¼ a4 sinðax þ bÞ
...
...
1
½cosðax þ bÞ þ i sinðax þ bÞ
f 0 ðD2 Þ
1
½cosðax þ bÞ þ i sinðax þ bÞ
¼
f ðD2 Þ
Equating real and imaginary parts, we have
1
1
cosðax þ bÞ ¼ x: 0 2
cosðax þ bÞ
2
f ðD Þ
½ f ðD ÞD2 ¼a2
We note in general that
ðD2 Þn sinðax þ bÞ
or
ð89Þ
¼ ða2 Þn sinðax þ bÞ:
Hence
f ðD2 Þ sinðax þ bÞ ¼ f ða2 Þ sinðax þ bÞ:
ð90Þ
Operating on both sides by f ðD1 2 Þ, we get
provided f 0 (–a2) 6¼ 0.
If f 0 (–a2) ¼ 0, then repeating the above process, we have
1
1
cosðax þ bÞ ¼ x2 00 2
cosðax þ bÞ
f ðD2 Þ
½ f ðD ÞD2 ¼a2
1
sinðax þ bÞ:
sinðax þ bÞ ¼ f ða Þ
f ðD2 Þ
2
Dividing both sides by f (–a2), we have
1
1
sinðax þ bÞ ¼
sinðax þ bÞ:
f ða2 Þ
f ðD2 Þ
Hence
1
1
sinðax þ bÞ ¼
sinðax þ bÞ;
2
f ðD Þ
f ða2 Þ
provided f (–a2) 6¼ 0.
Similarly,
1
1
cosðax þ bÞ ¼
cosðax þ bÞ;
2
f ðD Þ
f ða2 Þ
provided that f 0 (–a2) 6¼ 0, and
1
1
sinðax þ bÞ ¼ x 0 2
sinðax þ bÞ;
f ðD2 Þ
½ f ðD ÞD2 ¼a2
ð87Þ
provided f 00 (–a2) 6¼ 0
and
1
1
sinðax þ bÞ ¼ x2 00 2
sinðax þ bÞ
f ðD2 Þ
½ f ðD ÞD2 ¼a2
provided f 00 (–a2) 6¼ 0.
Case III. When F(x) ¼ xn, n
ð88Þ
provided f (–a2) 6¼ 0.
If f (–a2) ¼ 0, then (87) and (88) are not valid.
In such a situation, we proceed as follows:
By Euler’s formula
eiðaxþbÞ ¼ cosðax þ bÞ þ i sinðax þ bÞ:
Thus
1
1
eiðaxþbÞ ¼
½cosðax þ bÞ þ i sinðax þ bÞ
f ðD2 Þ
f ðD2 Þ
being positive integer
Since in this case
1
1
FðxÞ ¼
xn ;
P:I:
f ðDÞ
f ðDÞ
we make the coefficient of the leading term of f (D)
unity, take the denominator in numerator and then
expand by Binomial theorem. Operate the resulting
expansion on xn.
Case IV. When F(x) ¼ eax Q(x), where Q(x) is some
function of x
Let G is a function of x, we have
D½eax G ¼ eax DG þ a eax G ¼ eax ðD þ aÞG
D2 ½eax G ¼ eax ðD þ aÞ2 G
or, by 86,
...
1
1
...
iðaxþbÞ
x: 0 2 e
½cosðax þ bÞ þ isinðax þ bÞ
¼
Dn ½eax G ¼ eax ðD þ aÞn G
f ðD Þ
f ðD2 Þ
Ordinary Differential Equations
f ðDÞ½eax G ¼ eax f ðD þ aÞG
C:F: ¼ c1 ex=2 þ c2 e3x=2 :
1
, we get
Operating both sides by f ðDÞ
Since 2 is not a root of the auxiliary equation,
by (82), we have
1
½eax f ðD þ aÞG
f ðDÞ
Putting f (D þ a)G ¼ Q, we have G ¼
we have
1
1
Q¼
ðeax QÞ
eax :
f ðD þ aÞ
f ðDÞ
Q
f ðDþaÞ
P:I ¼
and so
1 2x
1
e2x
e ¼
e2x ¼
:
f ð2Þ
4ð4Þ þ 4ð2Þ 2
21
Hence the complete solution of the given equation is
y ¼ C:F: þ P:I ¼ c1 ex=2 þ c2 ex=2 þ
or
1
1
ðeax QðxÞÞ ¼ eax
Q
f ðDÞ
f ðD þ aÞ
ð91Þ
Case V. When F(x) ¼ x Q(x)
Solution. The symbolic form of the equation is
f ðDÞ ¼ ðD m1 ÞðD m2 Þ . . . ðD mn Þ:
Then, using partial fractions and Theorem 1.7,
we have
1
P:I: ¼
FðxÞ
f ðDÞ
1
FðxÞ
ðD m1 Þ ðD m2 Þ . . . ðD mn Þ
A1
A2
An
FðxÞ
þ
þ ... þ
D m1 D m2
D mn
Z
Z
¼ A1 em1 x FðxÞ em1 x þ A2 em2 x FðxÞ em2 x
¼
Z
þ . . . þ An e
mn x
e2x
:
21
EXAMPLE 1.91
2
3x
Solve ddx2y 5 dy
dx þ 6y ¼ e .
Resolving f (D) into linear factors, we have
¼
1.45
The roots of A.E. are m ¼ 12, – 32. Therefore, the
complementary function is
Hence
eax G ¼
n
FðxÞemn x :
ðD2 5D þ 6Þy ¼ e3x :
The auxiliary equation is
m2 5m þ 6 ¼ 0
or
ðm 3Þðm 2Þ 0:
Therefore m ¼ 2, 3. Then
C:F: ¼ c1 e3x þ c2 e2x :
Since 3 is a root of auxiliary equation, we use (85)
and get
P:I ¼ x
e3x
e3x
¼x
¼ xe3x :
ð3Þ
ð3 2Þ
Hence, the complete solution of the given equation is
y ¼ C:F: þ P:I: ¼ c1 e3x þ c2 e2x þ x e3x :
EXAMPLE 1.90
2
2x
Solve 4 ddxy2 þ 4 dy
dx 3y ¼ e .
Solution. The symbolic form of the given differential
equation is
ð4D þ 4D 3Þy ¼ e
2
and so the auxiliary equation is
4m2 þ 4m 3 ¼ 0:
2x
Remark 1.2. (a) In the above example, if we use (86),
then
eax
P:I: ¼ x d
dD f ðDÞ D¼a
¼x
e3x
e3x
¼ x e3x :
¼x
½2D 5D¼3
65
(b) We can also find the particular integral in the
above case by using Theorem 1.7. In fact, we have
1.46
n
Engineering Mathematics-II
1
1
FðxÞ ¼
e3x
f ðDÞ
ðD 3Þ ðD 2Þ
Z
1
2x
FðxÞ e2x dx
¼
e
D3
Z
1
1
¼
e2x e3x e2x dx ¼
e3x
D3
D3
Z
Z
3x
3x
3x
3x
¼e
e : e dx ¼ e
e0 dx ¼ x e3x :
P:I: ¼
EXAMPLE 1.92
Solve
d2y
dy
3 þ 2y ¼ cos hx:
dx2
dx
which yields m ¼ ±2. Hence
C:F: ¼ c1 e2x þ c2 e2x :
Now, using (82) and (87), we have
1
1
1
ðex þsin2xÞ ¼ 2
ex þ 2
sin2x
D2 4
D 4
D 4
ex
1
1
1
þ
sin2x ¼ ex sin2x:
¼ 2
1 4 44
3
8
P:I: ¼
Hence, the complete solution of the given differential equation is
1
1
y ¼ C:F: þ P:I: ¼ c1 e2x þ c2 e2x ex sin 2x:
3
8
Solution. The auxiliary equation is
m2 3m þ 2 ¼ 0;
EXAMPLE 1.94
3
2
Solve ddx3y þ ddx2y dy
dx – y ¼ cos 2x.
which yields m ¼ 1, 2. Therefore,
Solution. The auxiliary equation is
C:F: ¼ c1 e þ c2 e :
2x
m3 þ m2 m 1 ¼ 0;
ex þ ex
:
2
whose roots are 1, –1, –1. Therefore,
x
Now
cosh x ¼
C:F: ¼ c1 ex þ ðc2 þ c3 xÞex
Therefore,
Further,
1
1
ex þex
P:I: ¼
FðxÞ ¼
1
1
2
f ðDÞ
ðD1ÞðD2Þ
cos2x
P:I: ¼
FðxÞ ¼ 3
2
f ðDÞ
D þD D1
1
1
1
x 1
x
e þ : 2
e
¼ : 2
2 D 3Dþ2
2 D 3Dþ2
1
cos2x
¼
1
1
1
2
x 1
x
DD D2 D1
e þ
¼ 2
e
2 D 3Dþ2
2 1þ3þ2
1
1
1
1
cos2x
¼
¼ xd 2
ex þ ex since f ð1Þ ¼ 0
Dð4Þþð4ÞD1
2 dD ½D 3Dþ2D¼1
12
x
1
1
1
1
¼
ex þ ex
cos2x ¼ cos2x
¼
2 ½2D3D¼1
12
5D5
5ðDþ1Þ
x
1
¼ ex þ ex :
ðD1Þ
1
1
2
12
¼
cos2x ¼ ðD1Þ
cos2x
5ðD2 1Þ
5
41
Hence, the complete solution of the equation is
x
1
1
1
y ¼ C:F: þ P:I: ¼ c1 ex þ c2 e2x ex þ ex :
¼ ðD1Þ cos2x ¼ ðDcos2xcos2xÞ
2
12
25
25
EXAMPLE 1.93
1
1
2
¼ ð2sin2xcos2xÞ ¼ ð2sin2xþcos2xÞ:
Solve ddx2y 4y ¼ ex þ sin 2x.
25
25
Solution. The auxiliary equation for the given differential equation is
m2 4 ¼ 0;
Hence the complete solution is
y ¼ c1 ex þ ðc2 þ c2 xÞex 1
ð2 sin 2x þ cos 2xÞ:
25
Ordinary Differential Equations
Solution. The auxiliary equation is m – 4 ¼ 0 and
m ¼ ±2. Therefore,
C:F: ¼ c1 e2x þ c2 e2x :
2
C:F: ¼ c1 e2x þ c2 e2x :
Further,
1
1
x2
FðxÞ ¼ 2
f ðDÞ
D 4
1
1
2
¼
x2 ¼ 2 x
2
4D
4 1 D4
1
1
D2
1
¼
x2
4
4
P:I: ¼
Further,
x x 1
1
e e
xsinhx
¼
x
2
D2 4
D2 4
1
1
1
xex 2
xex
2
2 D 4
D 4
"
#
1 x
1
1
x
¼ e
xe
x
2
ðDþ1Þ2 4
ðD1Þ2 4
¼
1 x
1
1
xex 2
x
e
2 D2 þ2D3
D 2D3
"
#
1 x
1
1
x
xe
x
¼ e
D2
D2
2
3 1 2D
3 1þ 2D
3 3
3 3
" (
!)1
1 x
2D D2
þ
¼ e 1
x
6
3
3
!)1 #
(
2
2D
D
ex 1þ
x
3
3
1
2D
2D
¼ ex 1þ þ... xex 1 þ... x
6
3
3
1
2
2
¼ ex xþ ex x
6
3
3
x x x x x e e
2 e þe
¼
2
2
3
9
1
D2
þ ::: x2
1þ
4
4
1
1
¼ x2 D2 ðx2 Þ
4
16
1
1
1
1
¼ x2 ð 2Þ ¼ x2 :
4
16
4
8
¼
¼
x
2
¼ sinhx coshx:
3
9
Hence the solution is
y ¼ C:F: þ P:I:
x
2
¼ c1 e2x þ c2 e2x sin hx cos hx:
3
9
EXAMPLE 1.96
2
Solve ddx2y – 4y ¼ x2.
1.47
Solution. The auxiliary equation is m4 – 4 ¼ 0 and so
m ¼ ± 2. Therefore,
EXAMPLE 1.95
2
Solve ddx2y – 4y ¼ x sin hx.
P:I: ¼
n
Hence the complete solution is
1
1
y ¼ C:F: þ P:I: ¼ c1 e2x þ c2 e2x x2 :
4
8
EXAMPLE 1.97
2
Solve ddx2y þ 4y ¼ ex þ sin 3x þ x2.
Solution. The auxiliary equation is m2 þ 4 ¼ 0 and so
m ¼ ± 2i. Therefore,
C:F: ¼ c1 cos 2x þ c2 sin 2x:
Further,
P:I: ¼
1
1
ðex þ sin 3x þ x2 Þ
FðxÞ ¼ 2
f ðDÞ
D þ4
1
1
1
ex þ 2
sin 3x þ 2
x2
þ4
D þ4
D þ4
1
1 x
1
1
D2
sin 3x þ
1þ
¼ e þ
x2
4
5
9 þ 4
4
1 x 1
1
D2
þ . . . x2
1
¼ e sin 3x þ
4
5
5
4
¼
D2
1
1
x2
1
¼ ex sin 3x þ :2
4 16
5
5
1
1
1
1
¼ ex sin 3x þ x2 :
5
5
4
8
1.48
n
Engineering Mathematics-II
Hence the complete solution is
Further
y ¼ C:F: þ P:I:
P:I: ¼
1
1
1
1
¼ c1 cos 2x þ c2 sin 2x þ ex sin 3x þ x2 :
5
5
4
8
EXAMPLE 1.98
Solve
d2 y
dx2
x
2 dy
dx þ y ¼ xe sin x.
Solution. The auxiliary equation of the given differential equation is
m2 2m þ 1 ¼ 0;
which yields m ¼ 1, 1. Hence
C:F: ¼ ðc1 þ c2 xÞex :
1
½sin3xcos2x
D2 4D þ 3
1
1
½ 2sin3xcos2x
¼ 2
D 4D þ 3 2
1
1
½ ðsin5x þ sinxÞ
¼ 2
D 4D þ 3 2
1
1
1
1
sin5x þ : 2
sinx
¼ : 2
2 D 4D þ 3
2 D 4D þ 3
1
1
1
¼
sin5x þ
sinx
2 25 4D þ 3
1 4D þ 3
¼
1
1
1
sin5x þ
sinx
2 22 4D
2 4D
¼
1
1
1
sin5x þ
sinx
2 2ð11 þ 2DÞ
2ð1 2DÞ
¼
1
11 2D
1 þ 2D
sin5x þ
sinx
4 121 4D2
1 4D2
¼
1
11 2D
1 þ 2D
sin5x þ
sinx
4 121 4ð25Þ
1 4ð1Þ
The particular integral is
1
1
xex sin x
FðxÞ ¼
P:I: ¼
f ðDÞ
ðD 1Þ2
¼ ex
1
2
x sin x ¼ ex
1
x sin x
D2
ðD þ 1 1Þ
Z
1
1
x sin x dx ¼ ex ðx cos x þ sin xÞ
¼ ex
D
D
Z
¼ ex ðx cos x þ sin xÞ dx
¼ ex ½x sin x cos x cos x
¼ ex ðx sin x þ 2 cos xÞ:
Hence the complete solution is
y ¼ C:F: þ P:I:
¼ ðc1 þ c2 xÞex ex ðx sin x þ 2 cos xÞ:
EXAMPLE 1.99
Solve (D2 – 4D þ 3)y ¼ sin 3x cos 2x .
Solution. The auxiliary equation is
m 4m þ 3 ¼ 0;
2
which yields m ¼ 3, 1. Therefore,
C:F: ¼ c1 e3x þ c2 ex :
1 11 2D
1 þ 2D
sin5x þ
sinx
4
221
5
"
1
1
½11sin5x 2Dsin5x
¼ 4
221
#
1
þ ðsinx þ 2DsinxÞ
5
¼
1
11
10
1
2
sin5x þ
cos5x þ sinx þ cosx
4 221
221
5
5
11
10
1
1
sin5x þ
cos5x þ sinx þ cosx:
¼
884
884
20
10
Hence the complete solution is
11
10
y ¼ c1 e3x þ c2 ex sin 5x þ
cos 5x
884
884
1
1
þ sin x þ cos x:
20
10
¼
EXAMPLE 1.100
Solve (D2 þ 1)y ¼ cosec x.
Solution. The auxiliary equation is m2 þ 1 ¼ 0, which
yields m ¼ ±i. Thus,
C:F: ¼ c1 cos x þ c2 sin x:
Ordinary Differential Equations
Now
n
1.49
Therefore,
1
1
P:I: ¼
cosec x
FðxÞ ¼ 2
f ðDÞ
D þ1
C:F: ¼ c1 e2x þ c2 e3x :
Further
1
cosec x
¼
ðD þ iÞ ðD iÞ
P:I: ¼
1
FðxÞ
f ðDÞ
¼
1
1
1
cosec x
2i D i D þ i
¼
¼
1
1
1
cosec x cosec x :
2i D i
Dþi
¼e
ðcot x iÞ dx
¼ eix ðlog sin x ixÞ:
Similarly,
1
cosec x ¼ eix ðlog sin x þ ixÞ:
Dþi
1
P:I: ¼ ½eix ðlog sin x ixÞ eix ðlog sin x þ ixÞ
2i
ix
ix
e eix
e þ eix
¼ log sin x
x
2i
2
¼ ðlog sin xÞ sin x x cos x:
Hence the complete solution is
y ¼ c1 cos x þ c2 sin x þ sin x log sin x x cos x:
sin 2x
1
sin 2x
þD
1
¼ e2x
sin 2x
D4
Dþ4
sin 2x
¼ e2x : 2
D 4
Dþ4
sin 2x
¼ e2x
8
Z
Therefore,
ðD 2Þ2 þ 5ðD 2Þ þ 6
¼ e2x
But, by Theorem 1.7,
Z
1
cosec x ¼ eix cosec x eix dx
Di
Z
ix
¼e
cosec x ðcos x i sin xÞ dx
¼ eix
1
e2x sin 2x
þ 5D þ 6
1
2x
D2
D2
¼
e2x
½D sin 2x þ 4 sin 2x
8
¼
e2x
½2 cos 2x þ 4 sin 2x:
8
Hence the complete solution is
y ¼ C:F: þ P:I:
1
¼ c1 e2x þ c2 e3x e2x ½2 cos 2x þ 4 sin 2x:
8
EXAMPLE 1.102
3
Solve ddx2y þ y ¼ sin 3x cos2 2x.
Solution. The auxiliary equation of the given differential equation is
m3 þ 1 ¼ 0
EXAMPLE 1.101
2
2x
sin 2x.
Solve ddx2y þ 5 dy
dx þ 6y ¼ e
Solution. The auxiliary equation is m2 þ 5m þ 6 ¼ 0,
which yields
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5
25 24
m¼
¼ 2; 3:
2
or
ðm þ 1Þðm2 m þ 1Þ ¼ 0
pffiffi
and so m ¼ –1, 12 ± 23 i. Therefore,
C:F: ¼ c1 ex
þe
x=2
pffiffiffi
pffiffiffi
3
3
x þ c3 sin
x :
c2 cos
2
2
1.50
n
Engineering Mathematics-II
On the other hand,
which yields m = 1, 1, 4. Therefore,
1
1
C:F: ¼ ðc1 þ c2 xÞex þ c3 e2x :
sin3x ð1þcosxÞ
P:I: ¼ 3
D þ1
2
Further
1
1
1
sin3x
ð1þcosxÞ
¼ 3
x2 ex
P:I: ¼ 3
D þ1
2ðD3 þ1Þ
D 3D þ 2
1
1
1
sin3x
½1þcosx
¼
x2
¼ ex
3
DD2 þ1
2ð1þD3 Þ
ðD þ 1Þ 3ðD þ 1Þ þ 2
1
1
1
1
½1þcosx
sin3x
¼
x2
¼ ex 3
x2 ¼ ex 2 9Dþ1
2ð1þD3 Þ
2
D þ 3D
3D 1 þ D3
"
1
1
1
2 #
cosx
sin3x ð1þD3 Þ1 ð1Þ
¼
1
D
D
3
x
19D
2
2ð1þD Þ
¼e
1 þ
:: x2
3D2
3
3
1þ9D
1
sin3x ð1D3 þ::Þð1Þ
¼
2
ex
D D2
181D
2
. . . x2
¼
1 þ
2
1
3D
9
3
cosx
2þ2DD2
ex
Dx2 D2 x2
2
þ
¼
x 1þ9D
1
1
3D2
3
9
¼
sin3x cosx
181ð9Þ
2 22D
x
e
2x 2
¼
x2 þ
1þ9D
1 2þ2D
2
3D
3 9
sin3x cosx
¼
730
2 44D2
Z
Z
x Z
e
2
2
2
1
1 2þ2D
x
dx
¼
x
dx
þ
dx
ðsin3xþ27cos3xÞ cosx
¼
3D
3
9
730
2
8
1
1 1
1
ex x3 x2 2
þ x
ðsin3xþ27cos3xÞ cosxþ sinx
¼
¼
730
2 4
4
3D 3
3 9
Z 2
Z
1
1 1
x Z
3
e
x
x
2
ðsin3xþ27cos3xÞ ðcosxsinxÞ:
¼
¼
dx dx þ
x dx
730
2 4
3
3
3
9
Hence the complete solution is
pffiffiffi
pffiffiffi
3
3
x þ c3 sin
xÞ
y ¼ c1 ex þ ex=2 ðc2 cos
2
2
1
1
þ
ðsin 3x þ 27 cos 3xÞ 730
2
1
ðcos x sin xÞ :
4
EXAMPLE 1.103
3
2 x
Solve ddx3y 3 dy
dx þ 2y ¼ x e :
Solution. The symbolic form of the given equation is
ðD3 3D þ 2Þy ¼ x2 ex :
Its auxiliary equation is
m3 3mþ2 ¼ 0;
¼
ex x4 x3 x2
ex
þ
¼
½3x4 4x3 þ 4x2 3 12 9
9
108
¼
x2 ex 2
½3x 4x þ 4:
108
Hence complete solution is
y ¼ C:F: þ P:I:
¼ ðc1 þ c2 xÞex þ c3 e2x þ
x2 ex 2
½3x 4x þ 4:
108
EXAMPLE 1.104
Solve (D2 1)y = x sin 3x + cos x.
Solution. The auxiliary equation for the given differential equation is m2 1 = 0, and so m = ±1.
Therefore,
C:F: ¼ c1 ex þ c2 ex :
Ordinary Differential Equations
Further,
1
ðx sin3xþcosxÞ
P:I: ¼ 2
D 1
1
1
xðI:P: of e3ix Þ þ 2
cos x
¼ 2
D 1
D 1
1
1
x e3ix þ
cos x
¼ I:P: of 2
D 1
ð12 Þ 1
"
#
1
1
¼ I:P:of e3ix
x cos x
2
2
ðD þ 3iÞ 1
1
cos x
¼ I:P:of e3ix 2
x D þ 6iD 10
2
"
#
1
1
cos x
¼ I:P:of e3ix
2 x
6
2
10 1 10
iD D10
" #
1
2
1
3
D
iD þ
¼ I:P:of e3ix 1 x
10
10
5
1
cos x
2
1
3
D2
þ :: x
¼ I:P:of e3ix 1 þ iD þ
10
10
5
1
cos x
2
1
3
1
3ix
x þ i cos x
¼ I:P:of e
10
5
2
1
3
¼ I:P:of ðcos 3x þ i sin 3xÞ x þ i
10
5
1
cos x
2
1
3i
¼ I:P:of½xcos3xþixsin3xþ cos 3x
10
5
3
1
sin3x cos x
5
2
1
3
¼ I:P:of x cos 3x sin 3x
10
5
3
1
þi cos3x þ xsin3x cos x
5
2
1 3
1
cos 3x þ x sin 3x cos x:
¼
10 5
2
Hence the complete solution of the given differential equation is
y ¼ C:F: þ P:I: ¼c1 ex þc2 ex
1 3
1
cos 3x þ x sin 3x cos x:
10 5
2
1.17
n
1.51
METHOD OF VARIATION OF PARAMETERS TO
FIND PARTICULAR INTEGRAL
Definition 1.23. Let y1(x), y2(x), . . ., yn(x) be the
functions defined on [a, b] such that each function
possesses n 1 derivatives on [a, b]. Then the
determinant
y1
y2
:::
yn y0
y02
:::
y0n 1
:::
:::
::: W ðy1 ; y2 . . . ; yn Þ ¼ :::
:::
:::
:::
::: ðn1Þ
ðn1Þ
ðn1Þ y
y2
::: yn
1
is called the Wronskian of the set {y1, y2, . . ., yn}.
If the Wronskian of a set of n functions on [a, b]
is non-zero for atleast one point in [a, b], then the
set of n functions is linearly independent.
If the Wronskian is identically zero on [a, b] and
each of the function is a solution of the same linear
differential equation, then the set of functions is
linearly dependent.
The method of variation of parameters is applicable to the differential equation of the form
d2y
dy
þ p þ qy ¼ FðxÞ;
ð92Þ
2
dx
dx
where p, q, and F are functions of x.
Let the complementary function of (92) be
y=c1y1+c2y2.. Then y1 and y2 satisfy the equation
d2y
dy
þ p þ qy ¼ 0
ð93Þ
dx2
dx
Replacing c1 and c2 (regarded as parameters) by
unknown functions u(x) and v(x), we assume that
particular integral of (92) is
y ¼ uy1 þvy2 :
ð94Þ
Differentiating (94) with respect to x, we get
y0 ¼ uy01 þ xy02 þ u0 y1 þ x0 y2
¼ uy01 þ xy02 ;
under the assumption that
u 0 y1 þ v0 y2 ¼ 0
ð95Þ
ð96Þ
Differentiating (95) with respect to x, we get
y00 ¼ uy001 þvy002 þu0 y1 þv0 y02 :
ð97Þ
Substituting the values of y, y0 , and y 00 from (94),
(95), and (97) in (92), we have
uy1 00 þvy2 00 þu0 y1 0 þv0 y2 0 þpðuy1 0 þvy2 0 Þ
þ qðuy1 þvy2 Þ ¼FðxÞ
1.52
n
Engineering Mathematics-II
or
Therefore,
00
0
00
0
uðy1 þpy1 þ qy1 Þþvðy2 þpy2 þqy2 Þ
Since y1, y2 satisfy (93), the above expression
reduces to
ð98Þ
Solving (96) and (98), we get
u0 ¼ where
y
W ¼ 10
y1
y2 FðxÞ
y1 FðxÞ
and v0 ¼
;
W
W
y2 FðxÞ
dx þ y2
W
Z
y1 FðxÞ
dx
W
Z
Z 3x 3x
x e3x :e3x
e :e
3x
dx
þ
xe
¼ e3x
2
6x
x e
x2 e6x
Z
Z
dx
1
¼ e3x
dx
þ xe3x
x
x2
1
¼ e3x log x þ x e3x ¼ e3x ðlog x þ 1Þ:
x
P:I: ¼ y1
þ u0 y1 0 þv0 y2 0 ¼FðxÞ:
u0 y01 þv0 y02 ¼ FðxÞ:
Z
Hence the complete solution is
y ¼ C:F: þ P:I: ¼ ðc1 þ c2 xÞe3x e3x ðlog x þ 1Þ
y2 ¼ y1 y02 y1 0 y2 :
y02 ¼ ½kþc2 xlog xe3x where k ¼ c1 1:
Integrating, we have
Z
Z
y2 FðxÞ
y1 FðxÞ
dx ; v ¼
dx:
u¼
W
W
EXAMPLE 1.106
Using method of variation of parameters, solve
d2 y
dx2 þ y ¼ sec x.
Substituting the value of u and v in (94), we get
Z
Z
y2 FðxÞ
y1 FðxÞ
P:I: ¼ y ¼ y1
dx þ y2
dx:
W
W
Solution. The auxiliary equation for the given differential equation is m2 + 1 = 0 and so m = ± i. Thus
EXAMPLE 1.105
Using method of variation of parameters, solve
d2 y
dy
e3 x
dx2 6 dx þ 9y ¼ x2 .
To find P.I., let
Solution. The auxiliary equation for the given differential equation is
m2 6mþ9 ¼ 0;
Then
cos x sin x
W ¼ sin x cos x
¼ cos2 xþsin2 x¼ 1:
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
P:I: ¼ y1
W
W
Z
Z
sin x sec x
cos x sec x dx
dx þ sin x
¼ cos x
1
1
¼ cos x log cos xþx sin x:
C:F: ¼ ðc1 þc2 xÞ e3x .
Thus, we get
y1 ¼ e3x and y2 ¼ x e3x :
¼ e6x :
y1 ¼ cos x and y2 ¼ sin x:
Therefore,
which yields m = 3, 3. Therefore,
The Wronskian of y1, y2 is
y1 y2 W¼ 0
y1 y02 e3x
x e3x
¼
3e3x ð3x þ 1Þe3x
C:F: ¼ c1 cos xþc2 sin x:
Z
Hence the complete solution is
y ¼ C :F: þ P:I: ¼c1 cos xþc2 sin x
þ cos x log cos xþx sin x:
EXAMPLE 1.107
Solve the given equation using method of variation
of parameters
d2 y
dx2 þy¼ cosec x:
Ordinary Differential Equations
Solution. The symbolic form of the differential
equation is
ðD2 þ1Þy ¼ cosec x:
Its auxiliary equation is m2 + 1 = 0 and so m = ± i.
Therefore,
C:F: ¼ c1 cos xþc2 sin x:
To find P.I., let
y1 ¼ cos x
Then Wronskian
y y2
W ¼ 10
y1 y02
Therefore,
Z
P:I: ¼ y1
Z
¼ cos x
and
¼
y2 ¼ sin x:
cos x sin x sin x cos ¼ 1:
y2 FðxÞ
dx þ y2
W
Z
y1 FðxÞ
dx
W
sin x cosec x dx
Z
þ sin x
cos x cosec x dx
Z
Z
cos x
¼ cos x dx þsin x
dx
sin x
¼ x cos xþsin x log sin x:
Therefore,
n
1.53
Z
Z
y2 FðxÞ
y1 FðxÞ
P:I: ¼ y1
dx þ y2
dx
W
W
Z 2x x
Z x
e :e sin x
e sin x
¼
dx þ e2x
dx
2e2x
2e2x
Z
Z
1
e2x
ex sin x dx
e x sin x dx þ
¼
2
2
1
¼ e x sin x:
2
Hence the complete solution is
1
y ¼ C:F: þ P:I: ¼ c1 þ c2 e2x e x sin x:
2
EXAMPLE 1.109
Solve y 00 2y0 + 2y = e x tan x.
Solution. The auxiliary
equation is m2 2m + 2 = 0
pffiffiffiffiffiffi
and so m ¼ 2 248 ¼ 1 i: Hence,
C:F: ¼ex ðc1 cos xþc2 sin xÞ:
Let
y1 ¼ ex cos x and y2 ¼ ex sin x:
Then the Wronskian of y1, y2 is
ex sin x
ex cos x
W ¼ x
x
e ðcos x sin xÞ e ðcos x þ sin xÞ
¼ e2x :
Hence the complete solution is
y ¼ C :F: þ P:I: ¼c1 cos xþc2 sin x
x cos xþsinx log sinx:
EXAMPLE 1.108
Solve the given equation using method of variation
of parameters
d2y
dy
2 ¼ e x sin x:
2
dx
dx
Solution. The auxiliary equation is m2 2m = 0 or
m(m 2) = 0 and so m = 0,2. Hence
C:F: ¼ c1 þc2 e :
2x
Now let
y1 ¼ 1 and y2 ¼ e2x :
Then the Wronskian of y1, y2 is
1 e2x ¼ 2e2x :
W ¼
0 2e2x Therefore
Z
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
P:I: ¼ y1
W
W
Z x
e sin xex tan x
¼ ex cos x
dx
e2x
Z x
e cos xex tan x
dx
þ ex sin x
e2x
Z
¼ ex cos x ðsec x cos xÞ dx
Z
þ ex sin x sin x dx
¼ ex cos x½logðsec x þ tan xÞ sin x
ex sin x cos x
¼ ex cos x logðsec x þ tan xÞ:
Hence the complete solution is
y ¼ C :F: þ P:I: ¼ ex ðc1 cos x þ c2 sin xÞ
ex cos x logðsec x þ tan xÞ:
1.54
n
Engineering Mathematics-II
EXAMPLE 1.110
Using method of variation of parameters, solve the
differential equation
d2y
þ 4y¼ tan 2x:
dx2
Solution. The symbolic form of the given differential
equation is
ðD2 þ4Þy ¼ tan 2x:
Its auxiliary equation is m2 + 4 = 0, which yields m =
± 2i. Thus,
C:F: ¼ c1 cos 2xþc2 sin 2x:
To find P.I., let
y1 ¼ cos 2x and y2 ¼ sin 2x:
Then Wronskian W is
cos 2x
sin 2x
W ¼ 2 sin 2x 2 cos 2x
Hence,
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
W
W
Z
cos 2x
sin 2x tan 2x dx
¼
2
Z
sin 2x
þ
cos 2x tan 2x dx
x
Z
cos 2x sin2 2x
dx
¼
2
cos 2x
Z
sin 2x
þ
cos 2x tan 2x dx
2
Z
cos 2x 1 cos2 2x
dx
¼
2
cos 2x
Z
sin 2x
þ
cos 2x tan 2x dx
2
Z
1
¼ cos 2x ðsec 2x cos 2xÞdx
2
Z
1
þ sin 2x sin 2x dx
2
1
¼ cos 2x½logðsec 2x þ tan 2xÞ sin 2x
4
1
sin 2x cos 2x
4
1
¼ cos 2x logðsec 2x þ tan 2xÞ:
4
P:I: ¼ y1
Z
¼ 2:
Hence the complete solution is
y ¼ C:F: þ P:I: ¼c1 cos 2xþc2 sin2x
1
cos 2x logðsec 2x þ tan 2xÞ:
4
EXAMPLE 1.111
Solve the equation using the method of variation of
parameters
d2y
dy
2 þ y ¼ ex log x:
dx2
dx
Solution. The auxiliary equation is m2 2m + 1 = 0
and so m = 1, 1. Thus,
C:F: ¼ ðc1 þc2 xÞe x :
To find P.I., let y1 = ex, y2 = xex. Then
x
e
x ex
2x
W ¼ x
x ¼ e :
e ðx þ 1Þe
Therefore, Z
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
P:I: ¼ y1
W
W
Z
Z x x
x x
x e :e
e :e log x
¼ e x
log x dx þ xe x
dx
2x
e2x
Z e
Z
¼ e x x log x dx þ xe x log x dx
Z 2
x2
x
¼ e x
log x dx
2
2x
Z
x
dx
þ x e x x log x x
2
x
x2
log x þ x2 e x log x x2 e x
¼ e x
2
4
x2
x2
¼ e x log x þ e x þ x2 e x log x x2 e x
2
4
1 x 2
3 2 x 1 2 x
¼ e x log x x e ¼ x e ð2 log x 3Þ:
2
4
4
Hence the complete solution is
1
y ¼ C:F: þ P:I: ¼ ðc1 þc2 xÞe x þ x2 e x ð2 log x 3Þ:
4
1.18
DIFFERENTIAL EQUATIONS WITH VARIABLE
COEFFICIENTS
In this section, we shall consider differential
equations with variable coefficients which can be
reduced to linear differential equations with constant coefficients.
n
Ordinary Differential Equations
The general form of a linear equation of second
order is
d2y
dy
þ P þ Qy ¼ R;
dx2
dx
where P, Q and R are functions of x only. For
example,
d2y
dy
þ x þ x2 y ¼ cos x
dx2
dx
is a linear equation of second order.
If the coefficients P and Q are constants, then
such type of equation can be solved by finding
complementary function and particular integral as
discussed earlier. But, if P and Q are not constants
but variable, then there is no general method to
solve such problem. In this section, we discuss three
special methods to handle such problems.
(A) Method of Solution by Changing Independent
Variable
Let
d2y
dy
þ P þ Qy ¼ R;
ð99Þ
dx2
dx
be the given linear equation of second order. We
change the independent variable x to z by taking
z ¼ f ðxÞ. Then
dy dy dz
¼ : ;
dx dz dx
and
d 2 y dy d 2 z
d 2 y dz 2
þ
¼
:
dx2 dz dx2
dz2 dx
Substituting these values in (99), we get
d 2 y dz 2 dy d 2 z
dy dz
þP : þQy¼R
þ
2
2
dz dx
dz dx
dz dx
or
d 2 y dz
dz2 dx
2
or
d 2 y dy
þ
dz2 dz
þ
"
dy d 2 z
dz
þQy¼R
þp
dz dx2
dx
#
dz
þ P dx
Q
R
þ 2 y ¼ 2
dz2
dz
dz
d2 z
dx2
dx
dx
or
d2y
dy
þ P1 þ Q1 y ¼ R1 ;
dz2
dz
dx
1.55
where
d2 z
P1 ¼ dx
2
dz
þ P dx
;
dz 2
dx
Q
Q1 ¼
dz 2
dx
;
R1 ¼
R
:
dz 2
and
dx
Using the functional relation between z and x, it
follows that P1 ; Q1 and R1 are functions of x.
Choose z so that P1 ¼ 0, that is,
d2z
dz
þ P ¼ 0;
2
dx
dx
which yields
dz
¼ e
dx
or
Z
z¼
e
R
R
Pdx
Pdx
dx:
If for this value of z, Q1 becomes constant or a
constant divided by z2 , then the equation (99) can be
integrated to find its solution.
EXAMPLE 1.112
2
2
Solve ddx2y þ cot x dy
dx þ 4y cosec x ¼ 0.
Solution. Comparing with the standard form, we get
P ¼ cot x; Q ¼ 4 cosec2 x and
R ¼ 0:
If we choose z such that P1 ¼ 0, then z is given by
Z
R
Pdx
dx
z¼ e
Z
R
¼ e cot x dx dx
Z
¼ elog cosec x dx
Z
x
¼ cosec x dx ¼ log tan :
2
Then
Q
4 cosec2 x
¼ 4;
Q1 ¼ 2 ¼
dz
cosec2 x
dx
1.56
n
Engineering Mathematics-II
and
R
R1 ¼ 2 ¼
dz
dx
0
¼ 0:
cosec2 x
Therefore the given equation reduces to
d2y
þ 4y ¼ 0;
dz2
or, in symbolic form,
ðD2 þ 4Þy ¼ 0:
The auxiliary equation for this differential equation
is m2 þ 4 ¼ 0, which yields m ¼ 2i. Therefore the
solution of the given equation is
y ¼ c1 cos 2z þ c2 sin 2z
x
x
¼ c1 cos 2 log tan þ c2 sin 2 log tan :
2
2
EXAMPLE 1.113
2
3
5
Solve x ddx2y dy
dx 4x y ¼ x .
Solution. Dividing throughout by x, the given differential equation reduces to
d 2 y 1 dy
4x2 y ¼ x4 :
dx2 x dx
Comparing with the standard form, we have
1
P ¼ ; Q ¼ 4x2 ; and R ¼ x4 :
x
Choose z so that P1 ¼ 0. Then
Z
Z R
Z
R
1
Pdx
dx
x
z¼ e
dx ¼ e
dx ¼ elog x dx
Z
¼
x dx ¼
x2
:
2
Further,
Q
4x2
Q1 ¼ 2 ¼ 2 ¼ 4;
dz
x
dx
and
R
x4
R1 ¼ 2 ¼ 2 ¼ x2 :
dz
x
dx
Hence the given equation reduces to
d2y
4y ¼ x2 ¼ 2z
dx2
or
ðD2 4Þy ¼ 2z:
The auxiliary equation for this differential equation
is m2 4 ¼ 0 and so m ¼ 2. Therefore
C.F ¼ c1 e2z þ c2 e2z ¼ c1 ex þ c2 ex :
2
2
Now
1
1
2
D2
ð2zÞ
¼
1
ðzÞ
4
D2 4
4
1
D2
þ ::: ðzÞ
¼ 1þ
4
2
z
x2
¼ ¼ :
4
2
Hence the complete solution of the given differential equation is
1
2
2
y ¼ C.F þ P.I ¼ c1 ex þ c2 ex x2
4
P.I ¼
EXAMPLE 1.114
2
3
Solve x2 ddx2y 2x dy
dx þ 2y ¼ x .
(This equation is Cauchy–Euler equation and has
also been solved in Example 1.121 by taking
t ¼ log x).
Solution. Dividing the given equation throughout
by x2 , we get
d 2 y 2 dy 2
þ y ¼ x:
dx2 x dx x2
Comparing with the standard form, we have
2
2
P ¼ ; Q ¼ 2 and R ¼ x:
x
x
Choosing z such that P1 ¼ 0, we have
Z
Z
R
R1
Pdx
z¼ e
dx ¼ e2 xdx dx
Z
Z
x3
2 log x
dx ¼ x2 dx ¼ :
¼ e
3
Therefore the given equation reduces to
d2y
þ Q1 y ¼ R1 ;
dz2
where
2
Q
2
2
2
Q 1 ¼ 2 ¼ x 2 ¼ 6 ¼ 2
dz
x
9z
ðx2 Þ
dx
and
R1 ¼
R
dz 2
dx
¼
x
1
1
¼ ¼ :
x4 x3 3z
Ordinary Differential Equations
Hence the equation reduces to
n
1.57
Then
2
d y 2y
1
þ
¼
dx 9z2 3z
or
d2y
þ 2y ¼ 3z:
dz2
Let X ¼ log z so that z ¼ eX and (see Example
2
1.121), z2 ddz2y ¼ DðD 1Þy. Thus the equation now
reduces to
9z2
½9DðD 1Þ þ 2y ¼ 3eX
dy
dz
dv
¼v þz ;
dx
dx
dx
d2y
d2z
dv dz
d2v
:
þ
z
¼
v
þ
2
:
dx2
dx2
dx dx
dx2
Therefore equation (100) reduces to
d2v
dz
dv
z 2 þ 2 þ Pz
dx
dx
dx
2
d z
dz
þ
P
þ
Qz
v¼R
þ
dx2
dx
or
or
ð9D2 9D þ 2Þy ¼ 3eX :
The auxiliary equation for this symbolic equation is
m2 9m þ 2 ¼ 0;
which yields m ¼ 23 ; 13. Hence the complementary
function is given by
2
1
C.F ¼ c1 e3X þ c2 e3X
2
d2v
2 dz dv 1 d 2 z
dz
R
þ Pþ
þ P þ Qz v ¼
þ
dx2
z dx dx z dx2
dx
z
or
d2v
dv
þ P1 þ Q 1 v ¼ R 1 ;
2
dx
dx
where
2 dz
;
P1 ¼ P þ
z dx
2
1 d z
dz
R
Q1 ¼
þ P þ Qz ; R1 ¼ :
z dx2
dx
z
1
¼ c1 e3 log z þ c2 e3 log z
2
1
¼ c1 z3 þ c2 z3 ¼ c3 x2 þ c4 x:
Further,
1
eX
9D2 9D þ 2
1
3
3
eX ¼ eX ¼ elog z
¼3
99þ2
2
2
3
3 x3
¼ z¼
2
2 3
1 3
¼ x :
2
Hence the complete solution of the given differential equation is
1
y ¼ C.F þ P.I ¼ c3 x2 þ c4 x þ x3 :
2
(B) Method of Solution by Changing the
Dependent Variable
This method is also called the ‘‘Method of
Removing First Derivative’’. Let
P.I ¼ 3
By making proper choice of z, any desired value
can be assigned to P1 or Q1 . In particular, if P1 ¼ 0,
then
2 dz
Pþ : ¼0
z dx
or
dz
Pz
¼
dx
2
or
dz
1
¼ Pdx:
z
2
Integrating, we get
Z
1
Pdx
log z ¼ 2
or
z ¼ e 2
1
2
d y
dy
þ P þ Qy ¼ R
ð100Þ
dx2
dx
be the given linear equation of second order. We
change the dependent variable y by taking y ¼ vz.
ð101Þ
R
Pdx
Therefore,
dz
P 1
¼ e 2
dx
2
R
:
Pdx
1.58
n
Engineering Mathematics-II
and
R
1
2
d z
¼ e 2
dx2
Hence the required solution is
Pdx
y ¼ vz ¼ ðc1 e2x þ c2 e2x Þ sec x:
2
P
1 dP
:
4 2 dx
Putting these values in Q1 , we get
1 dP P2
Q1 ¼ Q :
4
2 dx
Hence the equation (101) reduces to
d2v
1 dP P2
v ¼ R1 :
þ Q
4
dx2
2 dx
EXAMPLE 1.116
2
2
x2
Solve ddx2y 4x dy
dx þ ð4x 3Þy ¼ e :
Solution. We have
1 dP P2
4
2 dx
1
2
¼ 4x 3 ð4Þ 4x2
2
¼ 1
Q1 ¼ Q ð102Þ
The equation (102) is called the normal form of the
given differential equation (100) and we observe
that this equation does not have first derivative.
That is why, the present method is called the
‘‘Method of Removing First Derivative’’. The
normal form (102) can be solved by using already
discussed methods.
EXAMPLE 1.115
2
Solve ddxy2 2 tan x dy
dx 5y ¼ 0 by the method of
removing first derivative.
Solution. Comparing the given equation with standard form, we have
P ¼ 2 tan x ; Q ¼ 5 ; R ¼ 0:
and
z ¼ e2
1
Pdx
¼ e2
R
x dx
2
¼ ex :
The normal form of the given equation is
2
2
d2v
ex
ex
¼ x2 ¼ 1
v¼
2
z
dx
e
or
ðD2 1Þv ¼ 1:
Therefore A.E is m2 1 ¼ 0, which yields m ¼
Thus,
C.F ¼ c1 ex þ c2 ex :
1.
Moreover,
Therefore,
P.I ¼ 1:
1 dP P2
¼ 5 þ sec2 x tan2 x
Q1 ¼ Q 4
2 dx
¼ 5 þ ð1 þ tan2 xÞ tan2 x ¼ 4;
and
z¼e
R
12
R
Pdx
¼e
R
Hence,
v ¼ C.F þ P.I ¼ c1 ex þ c2 ex 1:
But y ¼ vz. Therefore
y ¼ ðc1 ex þ c2 ex 1Þex :
2
tan x dx
¼ sec x:
The normal form of the given equation is
d2v
þ Q 1 v ¼ R1
dx2
or
d2v
4v ¼ 0
dx2
or
ðD2 4Þv ¼ 0:
The auxiliary equation for this symbolic form is
m2 4 ¼ 0. Therefore m ¼ 2 and so
v ¼ c1 e2x þ c2 e2x
(c) Method of Undetermined Coefficients
This method is used to find Particular integral of the
differential equation FðDÞ ¼ X , where the input
(forcing) function X consists of the sum of the
terms, each of which possesses a finite number of
essentially different derivatives. A trial solution
consisting of terms in X and their finite derivatives
is considered. Putting the values of the derivatives
of the trial solution in f ðDÞ and comparing the
coefficient on both sides of f ðDÞ ¼ X , the P I can
be found. Obviously the method fails if X consists
of terms like sec x and tan x having infinite number
of different derivatives. Further, if any term in
the trial solution is a part of the complimentary
Ordinary Differential Equations
n
1.59
function, then that term should be multiplied by x
and then tried.
which yields m ¼ 2; 2. Hence
EXAMPLE 1.117
2
Solve ddx2y þ y ¼ ex þ sin x.
The forcing function is ex sin x. Its derivative is
ex cos x þ ex sin x. Therefore, we consider
Solution. The symbolic form of the given differential
equation is
ðD2 þ 1Þy ¼ ex þ sin x:
The auxiliary equation is m þ 1 ¼ 0. Therefore
m ¼ i and so
C.F ¼ c1 cos x þ c2 sin x:
2
The forcing function consists of terms ex and sin x.
Their derivatives are ex and cos x. So consider the
trial solution.
y ¼ aex þ bx sin x þ c x cos x:
Then,
dy
¼ a ex þ bðx cos x þ sin xÞ þ cðcos x x sin xÞ
dx
d2y
¼ a ex þ b½ cos x x sin x þ cos x
dx2
þ c½ sin x sin x x cos x
¼ a ex þ 2b cos x bx sin x 2c sin x
c x cos x
x
¼ a e þ ð2b c xÞ cos x ðbx þ 2cÞ sin x:
2
Substituting the values of ddx2y and y in the given
equation, we get
2a ex þ 2b cos x 2c sin x ¼ ex þ sin x:
Comparing corresponding coefficients, we have
2a ¼ 1 ; 2b ¼ 0 and 2c ¼ 1:
Thus a ¼ 12 ; b ¼ 0 ; c ¼ 12 and so
1
1
P.I ¼ ex x cos x:
2
2
Hence the solution is
1
1
y ¼ c1 cos x þ c2 sin x þ ex x cos x:
2
2
EXAMPLE 1.118
2
x
Solve ddx2y 4 dy
dx þ 4y ¼ e sin x.
Solution. The symbolic form of the given differential
equation is
m2 4m þ 4 ¼ 0
C.F ¼ ðc1 þ c2 xÞe2x :
y ¼ aex sin x þ bex cos x
as the trial solution. Then
dy
¼ aðex cos x þ ex sin xÞ þ bðex cos x ex sin xÞ
dx
¼ ða þ bÞex cos x þ ða bÞex sin x;
d2y
¼ ða þ bÞ½ex cos x ex sin x
dx2
þ ða bÞ½ex cos x þ ex sin x
¼ 2a ex cos x 2bex sin x:
Substituting the values of
d 2 y dy
dx2 ; dx
and y in the
given differential equation, we get
2a ex cos x 2b ex sin x 4ða þ bÞex cos x
4ða bÞex sin x þ 4aex sin x þ 4bex cos x
¼ ez sin x
or
2aex cos x þ 2bex sin x ¼ ex sin x:
Comparing coefficients, we get 2b ¼ 1 or b ¼ 12.
Hence
1
P.I ¼ ex cos x
2
and so the complete solution of the given differential equation is
1
y ¼ C.F þ P.I ¼ ðc1 þ c2 xÞe2x þ ex cos x:
2
(D) Method of Reduction of Order
This method is used to find the complete solution of
d2 y
dy
dx2 þ P dx þ Q y ¼ R, where P, Q and R are function
of x only, and when part of complementary function
is known. So, let u, a function of x, be a part of the
complementary function of the above differential
equation. Then,
d2u
du
ð103Þ
þ P þ Q U ¼ 0:
dx2
dx
Let y ¼ u v be the complete solution of the given
differential equation, where v is also a function of x.
1.60
n
Engineering Mathematics-II
Then
dy
dv
du
¼u þv ;
dx
dx
dx
d2y
d2v
du dv
d2v
¼
u
þ
2
:
:
þ
v
dx2
dx2
dx dx
dx2
2
Substituting these values of ddxy2 ; dy
dx
and y in the given equation, we get
d2v
du dv
d2v
dv
du
þQuv
u 2 þ2 : þv 2 þP u þv
dx
dx dx
dx
dx
dx
Integrating (105) with respect to x, we get
Z
Z
R
R
1 Pdx
Pdx
v¼
R
u
e
e
dx þ c1 þ c2 ;
u2
where c1 and c2 are constants of integration. Hence
the complete solution of the given differential
equation is
y ¼ uv
¼u
Z
1 e
u2
R
Z
Pdx
R
Rue
Pdx
dx þ c1
þ c2 u:
¼R
or
2
d v
du
dv
d u
du
u 2 þ 2 þ Pu : þ
þ
P
þ
Q
u
v
dx
dx
dx
dx2
dx
2
¼R
or, using (103),
d2v
du
dv
¼R
u 2 þ 2 þ Pu
dx
dx
dx
2
Solution. Let y ¼ uv, where u ¼ ex be the complete
solution of the given equation. Then
or, division by u yields,
d2v
2 du
dv R
þ
þP
¼
2
dx
u dx
dx u
or, taking dv
dx ¼ z, we get
dz
2 du
R
þ
þP z¼ ;
dx
u dx
u
d2v
2 du dv
þ Pþ
¼ 0;
dx2
u dx dx
where
P ¼ 4x ; Q ¼ 4x2 2
ð104Þ
which is a first order differential equation in z and x.
The integration factor for (104) is
R 2 du
I:F ¼ e ðu dxþPÞdx
R
R 2
¼ e udxþ pdx
R
¼ u2 e pdx :
Therefore, the solution of (104) is
Z
R
R
R 2 pdx u e
z u2 e Pdx ¼
dx þ c1
u
or
1
R
z¼
u2 e
or
dv
1
¼ 2 e
dx u
R
Z
Pdx
R
R 2
u e
u
Z
Pdx
R
Ru e
Pdx
Pdx
and R ¼ 0.
Thus
d2v
2
dv
2
¼0
þ 4x þ x2 ð2xex Þ
dx2
dx
e
or
d2v
dv
þ ½4x 4x ¼ 0
dx2
dx
or
d2v
¼ 0:
dx2
Integrating, we get
dv
¼ c1 :
dx
Integrating once more, we get
v ¼ c1 x þ c 2 :
dx þ e
dx þ c1 :
EXAMPLE 1.119
Find the complete solution of
d 2y
dy
4x þ ð4x2 2Þy ¼ 0;
2
dx
dx
2
if y ¼ ex is an integral included in the complementary solution.
ð105Þ
Hence the complete solution is
2
y ¼ uv ¼ ex ½c1 x þ c2 :
Ordinary Differential Equations
n
1.61
EXAMPLE 1.120
2
Solve sin2 x ddx2y ¼ 2y, given that y ¼ cot x is an
integral included in the complementary function.
Hence the complete solution is
Solution. The given equation is
d2y
ð2cosec2 xÞy ¼ 0:
dx2
Comparing with
d2y
dy
þ P þ Q y ¼ R;
dx2
dx
we get
(E) Cauchy–Euler Homogeneous Linear Equation
Consider the following differential equation with
variable coefficients:
dny
d n1 y
dy
xn n þ a1 xn1 n1 þ ::: þ an1 x
dx
dx
dx
ð109Þ
þ an ¼ FðxÞ;
P ¼ 0; Q ¼ 2 cosec2 x; R ¼ 0:
Therefore, putting y ¼ uv ¼ ðcot xÞv, the reduced
equation is
d2v
2 du dv
¼ 0;
þ Pþ
dx2
u dx dx
y ¼ u v ¼ cot x½c1 ðtan x xÞ þ c2 ¼ c1 ð1 x cot xÞ þ c2 cot x:
where ai are constant and F is a function of x. This
equation is known as Cauchy–Euler homogeneous
linear equation (or equidimensional equation). The
Cauchy–Euler homogeneous linear equation can be
reduced to linear differential equation with constant
coefficients by putting x = et or t = log x. Then,
dy dy dt dy 1
¼ : ¼ :
dx dt dx dt x
that is,
d2v
2
dv
ðcosec2 xÞ
¼0
þ
dx2
cot x
dx
or
or
x
d2v
dv
2cosec2 x ¼ 0:
dx2
dx
dv
Let dx ¼ z. Then (106) reduces to
dz
cot x ð2cosec2 xÞz ¼ 0
dx
or
dz
cosec2 x
¼2
dx:
z
cot x
Integrating (107), we get
log z ¼ 2 log cot x þ log c1
cot x
ð106Þ
Now,
ð107Þ
dy dy
¼
¼ Dy:
dx dt
d2y
d 1 dy
1 dy 1 d 2 y dt
¼
:
¼
þ
dx2 dx x dt
x2 dt x dt2 dx
1 dy 1 d 2 y
¼ 2 þ 2 2
x dt x dt
and so
x2
d 2 y d 2 y dy
¼ D2 yDy¼ DðD 1Þy:
¼
dx2 dt2 dt
or
Similarly,
log z þ log cot2 x ¼ log c1
or
x3
z ¼ c1 tan2 x
or
dv
¼ c1 tan2 x ¼ c1 ðsec2 x 1Þ
dx
Integrating (108), we get
v ¼ c1 ðtan x xÞ þ c2 ;
where c1 and c2 are constants of integration.
ð108Þ
d3y
¼ DðD 1ÞðD 2Þy
dx2
and so on. Putting these values in (109), we obtain a
linear differential equation with constant coefficients which can be solved by using the methods discussed already.
EXAMPLE 1.121
2
3
Solve x2 ddx2y 2x dy
dx þ 2y ¼ x :
1.62
n
Engineering Mathematics-II
Solution. This is a Cauchy–Euler equation. Putting
2 d2 y
x = et or t = log x, we have x dy
dx ¼ Dy; x dx2 ¼
DðD 1Þy: Hence the given equation transforms to
Now
P:I: ¼
ðDðD 1Þ 2D þ 2Þy ¼ e3t
or
ðD2 3D þ 2Þy ¼ e3t ;
which is a linear differential equation with constant
coefficient. The auxiliary equation is m2 3m + 2 =
0 and so m = 1, 2. Therefore
C:F: ¼ c1 et þ c2 e2t :
The particular integral is
1
1
P:I: ¼
e3t
FðxÞ ¼ 2
f ðDÞ
D 3D þ 2
1
1
¼
e3t ¼ e3t :
99þ2
2
1 3t
e
2
Returning back to the variable x, we have
1
y ¼ c1 xþc2 x2 þ x3 :
2
EXAMPLE 1.122
Solve the Cauchy–Euler equation
d2y
dy
x2 2 x þ y ¼ log x:
dx
dx
Solution. Putting x = et, we have
dy
d2y
x ¼ Dy and x2 2 ¼ DðD 1Þy
dx
dx
and so the equation transforms to
ðDðD 1Þ D þ 1Þy ¼ t
or
ðD 1Þ2 y ¼ t:
The complementary function is
C:F: ¼ ðc1 þc2 tÞet :
t ¼ ð1 DÞ2 t
¼ ð1 þ 2D þ . . .Þt ¼ tþ2:
Hence the complete solution is
y ¼ C:F: þ P:I: ¼ ðc1 þc2 tÞet þtþ2:
Returning back to x, we get
y ¼ ðc1 þc2 log xÞxþlog xþ2:
EXAMPLE 1.123
Solve Cauchy–Euler equation
d2y
dy
x2 2 þ x þ y ¼ log x sinðlog xÞ:
dx
dx
Solution. Putting x = et, this equation transforms to
ðD2 þ1Þy ¼ tsin t:
Hence the complete solution is
y ¼ c1 et þ c2 e2t þ
1
ðD 1Þ2
The complementary function is
C:F: ¼ c1 cos tþc2 sin t:
Further
1
1
t sin t¼ I:P: of 2
teit
þ1
D þ1
1
¼ I:P: of ei t
ðD þ iÞ2 þ 1
1
t
¼ I:P: of ei t
D
2iD 1 þ 2i
1 1
iD 1
t
1
¼ I:P: of eit : :
2i D
2
it2 t
¼ I:P: of eit þ
4 4
2
it
t
þ
¼ I:P: of ðcos tþisin tÞ
4
4
2
t
t
¼ cos t þ sin t:
4
4
P:I: ¼
D2
Therefore, the complete solution is
t2
t
y ¼ c1 cos tþc2 sin t cos tþ sin t:
4
4
Ordinary Differential Equations
n
1.63
Returning back to x, we get
Moreover,
1
y ¼ c1 cosðlogxÞþc2 sinðlogxÞ ðlogxÞ2 cosðlogxÞ
4
1
þ log x sinðlog xÞ:
4
1
1
ðe2t þ t Þ
D2 D 2
e
1
1
¼ 2
e2t þ 2
ðet Þ
D D2
D D2
1
1
e2t þ t
et
¼t
½2D 1D¼2
ð2D 1ÞD¼1
t
1
¼ e2t tet :
3
3
Thus the complete solution is
t
t
y ¼ C :F: þ P:I: ¼c1 e2t þ c2 et þ e2t et
3
3
c
1
1
2
x2 log x:
¼ c1 x2 þ þ
x 3
x
EXAMPLE 1.124
Solve the Cauchy–Euler equation
x2
d2y
dy
x þ 2y ¼ x log x:
dx2
dx
Solution. Putting x = et, the equation reduces to
ðDðD 1Þ D þ 2Þy ¼ t et
or
ðD2 2D þ 2Þy ¼ t et
The C.F. for this equation is
C:F: ¼ et ðc1 cos tþc2 sin tÞ:
The particular integral is
1
t et
P:I: ¼ 2
D 2D þ 2
1
¼ et
t
2
ðD þ 1Þ 2ðD þ 1Þ þ 2
1
t ¼ et ð1 D2 Þ1 t ¼ et ðt0Þ ¼t et :
¼ et 2
D þ1
P:I: ¼
EXAMPLE 1.126
Solve the Cauchy–Euler equation
d2y
dy
x2 2 þ 2x 20y ¼ ðx þ 1Þ2 :
dx
dx
Solution. Putting x = et, the given equation transforms into
ðDðD 1Þ þ 2D 20Þy ¼ e2t þ 2et þ 1
or
ðD2 þD 20Þy ¼ e2t þ 2et þ 1:
The auxiliary equation is m2 + m 20 = 0 and so
m = 5, 4. Therefore,
C:F: ¼ c1 e5t þ c2 e4t :
Therefore, the complete solution is
y ¼ C :F: þ P:I: ¼ et ðc1 cos tþc2 sin tÞþt et
¼ xðc1 cosðlog xÞ
þ c2 sinðlog xÞþx log x:
EXAMPLE 1.125
2
Solve x2 ddx2y 2y ¼ x2 þ 1x :
Solution. Putting x = et, the given equation reduces to
ðDðD 1Þ 2Þy ¼ e2t þ e1t :
The auxiliary equation is m2 m 2 = 0 and so
m = 2, 1. Therefore,
C:F: ¼ c1 e2t þ c2 et :
Now,
P:I: ¼
1
2
e2t þ 2
et
D2 þ D 20
D D 20
1
x0 t
D2 D 20
1
2 t
1
¼ e2t þ
e :
14
20
20
þ
Thus the complete solution is
1 2t 1 t 1
e e 14
10
20
1
x
1
¼ c1 x5 þ c2 x4 x2 14
10 20
y¼ C:F: þ P:I: ¼c1 e5t þ c2 e4t 1.64
n
Engineering Mathematics-II
(F) Legendre’s Linear Equation
An equation of the form
dny
d n1 y
ðax þ bÞn n þ a1 ðax þ bÞn1 n1 þ :::
dx
dx
ð110Þ
þ an y¼ FðxÞ
where an are constants and F is a function of x, is
called Legendre’s linear equation.
To reduce the Legendre’s equation to a linear
differential equation with constant coefficient, we
put ax + b = et or t = log (ax + b). Then
dy dy dt
a dy
¼ : ¼
dx dt dx ax þ b dt
or
ðaxþbÞ
dy
¼ aDy:
dx
Further
2
d2y
d
a dy
a2
d y dy
¼
¼
dx2 dx ax þ b dt
ðax þ bÞ2 dt2 dt
and so
ðaxþbÞ2
d2y
¼ a2 ðD2 DÞy ¼ a2 DðD 1Þy:
dx2
Similarly,
ðaxþbÞ3
d3y
¼ a3 DðD 1ÞðD 2Þy
dx3
and so on.
Putting these values in (110), we get a linear
differential equation with constant coefficients
which can be solved by usual methods.
EXAMPLE 1.127
2
Solve ð2xþ3Þ2 ddx2y 2ð2x þ 3Þ dy
dx 12y ¼ 6x:
Solution. Putting 2x + 3 = et or t = log (2x + 3), the
given equation reduces to
ð4ðD2 DÞ 4D 12Þy ¼ 6
e 3
¼ 3et 9
2
The auxiliary equation is
4m2 8m12 ¼ 0;
which yields m = 3, 1. Therefore,
C:F: ¼ c1 et þ c2 e3t :
Now
P:I: ¼
4D2
1
3 t 3
ð3et 9Þ ¼
e þ :
8D 12
16
4
Hence the complete solution is
3
3
y ¼ c1 et þ c2 e3t et þ
16
4
¼
c1
3
3
þ c2 ð2x þ 3Þ3 ð2x þ 3Þ þ :
ð2x þ 3Þ
16
4
EXAMPLE 1.128
Solve the Legendre’s equation
ð1 þ xÞ2
d2y
dy
þ ð1 þ xÞ þ y ¼ 4 cos logð1 þ xÞ:
dx2
dx
Solution. Putting x + 1 = et or t = log (x + 1), the given
equations transforms to (12 (D2 D) + 1 D + 1)y= 4
cos t
or
ðD2 þ1Þy ¼ 4cos t:
The auxiliary equation is m2 + 1 = 0 and so m = ± i.
Therefore,
C:F: ¼ c1 cos tþc2 sin t:
The particular integral is
P:I: ¼
D2
4
4ð2DÞ
cos t ¼ t
cos t
þ1
½4D2 D2 ¼1
8t
D cos t ¼ 2tð sin tÞ ¼ 2t sin t:
4
Thus the complete solution is
¼
y ¼ c1 cos tþc2 sin tþ2tsin t
¼ c1 cosðlogðxþ1ÞÞþc2 sinðlogðxþ1ÞÞ
þ2logðx þ 1Þsinðlogðxþ1ÞÞ:
t
or
ð4D2 8D 12Þy ¼ 3et 9:
EXAMPLE 1.129
Solve the Legendre’s equation
d2y
dy
ð1þxÞ2 2 þ ð1 þ xÞ þ y¼ sinð2logð1þxÞÞ:
dx
dx
Solution. As in Example 1.128, putting x + 1 = et, the
given equation reduces to
ðD2 þ1Þy ¼ sinð2tÞ:
The complementary function is
C:F: ¼ c1 cos tþc2 sin t
Further,
P:I: ¼
1
1
1
sin½2t ¼
sin 2t ¼ sin 2t:
D2 þ 1
4 þ 1
3
Therefore, the complete solution is
1
y ¼ c1 cos tþc2 sin t sin2 t
3
¼ c1 cosðlogðxþ1ÞÞþc2 sinðlogðxþ1Þ
1
sinð2logðxþ1ÞÞ:
3
EXAMPLE 1.130
2
Solve (1 + x)2 ddx2y þ ð1 þ xÞ dy
dx þ y = 2 sin (log (x + 1)).
Solution. As in Example 1.128, putting x + 1 = et, the
given equation transforms to
ðD2 þ1Þy ¼ 2 sin t:
Its complementary function is given by
C:F: ¼ c1 cos tþc2 sin t:
Its particular integral is given by
1
2
ð2 sin tÞ ¼ 2
sin t
þ1
D þ1
2t
D sin t
¼ t cos t:
sin t ¼
¼
½2DD2 ¼1
1
P:I: ¼
D2
Therefore, the complete solution is
y ¼ c1 cos tþc2 sin ttcos t
¼ c1 cosðlogðxþ1ÞÞþc2 sinðlogðxþ1ÞÞ
logðxþ1Þcosðlogðxþ1ÞÞ:
1.19
SIMULTANEOUS LINEAR DIFFERENTIAL
EQUATIONS WITH CONSTANT COEFFICIENTS
Differential equations, in which there are two or
more dependent variables and a single independent
variable are called simultaneous linear equations.
The aim of this section is to solve a system of linear
differential equations with constant coefficients.
1.65
n
Ordinary Differential Equations
The solution is obtained by eliminating all but one
of the dependent variables and then solving the
resultant equations by usual methods.
EXAMPLE 1.131
Solve the simultaneous equations
dx
dy
¼ 7xy; ¼ 2xþ5y:
dt
dt
Solution. In symbolic form, we have
ðD 7Þxþy ¼ 0
ð111Þ
ðD 5Þy2x ¼ 0
ð112Þ
Multiplying (111) by (D 5) and subtracting (112)
from it, we get
ðD 5ÞðD 7Þxþ2x ¼ 0
or
ðD2 12D þ 35 þ 2Þx ¼ 0
or
ðD2 12D þ 37Þx ¼ 0:
Its auxiliary equation is m2 12m + 37 = 0, which
yields m = 6 ± i. Therefore, its complete solution is
x ¼ e6t ðc1 cos t þ c2 sin tÞ:
Now
dx d
d
¼ ðc1 e6t cos tÞ þ ðc2 e6t sin tÞ
dt dt
dt
¼ c1 ½6e6t cos t e6t sin t þ c2 ½6e6t sin t þ e6t cos t
¼ e6t ½ð6c1 þ c2 Þ cos t þ ð6c2 c1 Þ sin t
Putting the values of x and
dx
dt
in (101), we get
e ½ð6c1 þc2 Þ cos tþð6c2 c1 Þ sin t
6t
7e6t ½c1 cos tþc2 sin tþy¼ 0:
Therefore,
y ¼ 7e6t ðc1 cos tþc2 sin tÞ
e6t ½ð6c1 þc2 Þ cos tþð6c2 c1 Þ sin t
¼ e6t ½ðc1 c2 Þ cos tþðc1 þc2 Þ sin t:
Hence the solution is
x ¼ e6t ðc1 cos tþc2 sin tÞ
y ¼ e6t ½ðc1 c2 Þ cos tþðc1 þc2 Þ sin t:
EXAMPLE 1.132
2
Solve ddt2y þ dy
dt 2y ¼ sin t;
dx
dy þx3y¼
0:
1.66
n
Engineering Mathematics-II
Solution. The given system of equations are
or
ðD þD 2Þy ¼ sin t
ð113Þ
4D3 x þ 6D2 y ¼ 2Dð4Þ ¼ 0
ðD þ 1Þx 3y ¼ 0
ð114Þ
9Dx þ 6D2 y ¼ 0:
2
The auxiliary equation for (113) is
m2 þ m 2 ¼ 0
and so m = 2, 1. Therefore,
C:F: ¼ c1 et þ c2 e2t :
1
1
sin t ¼
sin t
P:I: ¼ 2
D þ D2 2
D3
Dþ3
1
sin t ¼ ðD þ 3sintÞ
¼ 2
D 9
10
3
1
¼ sin t cos t:
10
10
Therefore, the complete solution of (113) is
1
y ¼ c1 et þ c2 e2t ðcos t þ 3 sin tÞ:
10
Putting this value of y in (114), we get
dx
1
þx ¼ 3½c1 et þ c2 e2t ðcos t þ 3 sin tÞ
dt
R 10
The integrating factor is e 1 dt ¼ et . Therefore,
Z
1
x:et ¼ 3 et ½c1 et þ c2 e2t ðcos t þ 3 sin tÞdt;
10
which yields
3 t
x ¼ 32 c1 e2t 3e2 et 10
e ðcos t 2 sin tÞ.
Therefore, the solution is
3
3
x ¼ c1 e2t 3c2 et et ðcos t 2 sin tÞ
2
10
1
y ¼ c1 et þ c2 e2t ðcos t þ 3 sin tÞ:
10
EXAMPLE 1.133
A mechanical system with two degrees of freedom
satisfies the equation
d2x
dy
d2y
dx
2 2 þ 3 ¼ 4; 2 2 3 ¼ 0
dt
dt
dt
dt
dy
;
all
vanish at t = 0.
under the condition that x, y, dx
dt dt
Find x and y.
Solution. The given equations are
2D2 x þ 3Dy ¼ 4
2D2 y 3Dx ¼ 0
ð115Þ
Subtracting, we get
4D3 xþ9Dx¼ 0
or
ð4D3 þ9DÞx¼ 0:
Auxiliary equation is
4m3 þ 9m ¼ 0; and so m ¼ 0;
3
i:
2
Hence
3
3
x ¼ c1 þ c2 cos t þ c3 sin t:
2
2
At t = 0, x = 0. Therefore, 0 = c1 + c2 or c1 = c2.
Also
dx
3
3
3
3
¼ c2 sin t þ c3 cos t
dt
2
2
2
2
At t = 0,
dx
dt
= 0 and so 0 = 32 c3 or c3 = 0. Thus
3
x ¼ c1 c1 cos t:
2
Therefore,
dx
3
3
d2x 9
3
¼ c1 sin t and 2 ¼ c1 cos t:
dt
2
2
dt
4
2
Putting this value of dx
dt in (115), we get
9
3
9
3
2D2 y c1 sin t¼ 0 or D2 y ¼ c1 sin t:
2
2
4
2
Integrating, we get
3
3
Dy¼ c1 cos t þ k:
2
2
3
Using initial condition dy
dt = 0 at t = 0, we get k = 2c1.
Therefore,
3
3 3
Dy¼ c1 cos tþ c1 :
2
2 2
Integrating again, we have
3
3
y ¼ c1 sin t þ c1 t þ k:
2
2
When t = 0, y = 0. So we have k = 0. Hence
3t 3
y ¼ c1 sin þ c1 t:
2 2
Ordinary Differential Equations
2
Further, putting the value of ddt2x and dy
dt in the first of
8
9.
the given equations, we get c1 = Hence
8
3t
4
8
3t
1 cos
; y ¼ t sin :
x ¼
9
2
3
9
2
APPLICATIONS OF LINEAR DIFFERENTIAL
EQUATIONS
Linear differential equations play an important role
in the analysis of electrical, mechanical and other
linear systems. Some of the applications of these
equations are discussed below.
(A) Electrical Circuits
Consider an LCR circuit consisting of inductance L,
capacitor C, and resistance R. Then, using Kirchhoff’s
law [see (16)], the equation governing the flow of current in the circuit is
dI
Q
L þ RI þ ¼ EðtÞ;
dt
C
where I is the current flowing in the circuit, Q is
the charge, and E is the e.m.f. of the battery. Since
I = dQ
dt , the above equation reduces to
d2Q
dQ Q
L
þR
þ ¼ EðtÞ
ð116Þ
dt
dt C
If we consider LC circuit without having e.m.f.
source, then the differential equation describing
the circuit is
d2Q Q
L 2 þ ¼0
dt
C
or
d2Q Q
¼0
þ
dt2
LC
or
d2Q
1
þ m2 Q¼ 0;m2 ¼
:
dt2
LC
This equation represents free electrical oscillations
of the current having period
pffiffiffiffiffiffiffi
2
¼ 2 LC :
T¼
m
EXAMPLE 1.134
In an LCR circuit, an inductance L of one henry,
resistance of 6 ohm, and a condenser of 1/9 farad
have been connected through a battery of e.m.f.
1.67
E = sin t. If I = Q = 0 at t = 0, find charge Q and
current I.
Solution. The differential equation for the given circuit is
L
1.20
n
d2Q
dQ Q
þ ¼ EðtÞ:
þR
dt2
dt C
Here L = 1, R = 6, C = 19, E(t) = sin t. Thus, we have
d2Q
dQ
þ 9Q¼ sint
þ6
dt2
dt
subject to Q(0) = 0, Q0 (0) = 0 = I(0). The auxiliary
equation for this differential equation is m2 + 6m + 9
= 0 and so m = 3, 3. Thus
C:F: ¼ðc1 þ c2 tÞe3t :
Now
1
1
sin t ¼
sin t
D2 þ 6D þ 9
6D þ 8
6D 8
6D 8
sin t ¼
sin t
¼
2
36D 64
100
1
¼
½6D sin t 8 sin t
100
6
8
cos t þ
sin t:
¼
100
100
P:I: ¼
Hence the complete solution is
Q ¼ðc1 þ c2 tÞe3t 6
8
cos t þ
sin t:
100
100
6
6
Now Q(0) = 0 gives 0 = c1 100
and so c1 = 100
. Also
dQ
¼ 3c1 e3t þ c2 ðe3t 3te3t Þ
dt
6
8
þ
sin t þ
cos t:
100
100
Therefore dQ
dt = 0 at t = 0 yields
0 ¼ 3c1 þ c2 þ
8
18
8
¼
þ c2 þ
100
100
100
1
and so c2 = 10
. Hence
6
t
6
8
þ
e3t cos t þ
sin t
Q¼
100 10
100
100
e3t
3
2
ð5t þ 3Þ cos t þ sin t:
¼
50
50
25
1.68
n
Engineering Mathematics-II
dQ
dt ,
3t
Since I =
we have
5e
3
3
2
ð5t þ 3Þe3t þ sin t þ cos t
50
50
50
25
e3t
3
2
¼
ð15t þ 4Þ þ sin t þ cos t:
50
50
25
I¼
EXAMPLE 1.135
Find the frequency of free vibrations in a closed
electrical circuit with inductance L and capacity C
in series.
Solution. Since there is no applied e.m.f., the differential equation governing this LC circuit is
d2Q Q
L 2 þ ¼0
dt
C
or
d2Q
Q
¼ v2 Q;
¼
dt2
LC
1
where v2 = LC
. Thus the equation represents oscillatory current with period.
pffiffiffiffiffiffiffi
2
T¼
¼ 2 LC :
v
Then
ffi per second
Frequency = T1 ¼ 2p1 ffiffiffiffi
LC
60
30
¼ pffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffi per minute:
2 LC LC
EXAMPLE 1.136
The differential equation for a circuit in which
self-inductance and capacitance neutralize each
other is
d2i i
L 2 þ ¼ 0:
dt
C
Find the current i as a function of t, given that I is
maximum current and i = 0 when t = 0.
Solution. We have
d2i
i
¼ 0:
þ
dt2 LC
2
1
The auxiliary equation is m + LC
= 0 and so m =
i
ffi. Hence the solution is
± pffiffiffiffi
LC
1
1
i ¼ c1 cos pffiffiffiffiffiffiffi t þ c2 sin pffiffiffiffiffiffiffi t:
LC
LC
Since i = 0 at t = 0, we have c1 = 0 and so
t
i ¼ c2 sin pffiffiffiffiffiffiffi :
LC
t ffi
For maximum current I, we have I = c2 max sin pffiffiffiffi
LC
t ffi
ffiffiffiffi
p
= c2. Hence i = I sin LC .
EXAMPLE 1.137
An LCR circuit with battery e.m.f. E sin pt is tuned
1
. Show that for small
to resonance so that p2 = LC
R
value of L, the current in the circuit at time t is given
by E2Lt sin pt.
Solution. The differential equation governing the LCR
2
q
circuit is L ddt2q þ R dq
dt þ C ¼ EðtÞ ¼ E sin pt:
The auxiliary equation is
R
1
¼ 0;
m2 þ m þ
L
LC
which yields
qffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2
4
RL
L2 LC
m¼
2 ffiffiffiffiffiffiffiffiffi
q
R
4
L
LC
R
¼
is small
since
L
2
R
1
pffiffiffiffiffiffiffi i
¼
2L
LC
R
1
¼
pi since p2 ¼
:
2L
LC
Therefore,
C:F: ¼ e2L ðc1 cos pt þ c2 sin ptÞ
Rt
ðc1 cos ptþc2 sin ptÞ
¼ 1
2L
R
rejecting higher power of :
L
Rt
Further,
P:I: ¼
¼
¼
1
ðE sin ptÞ
LD2 þ RD þ C1
E
sin pt
Lp2 þ RD þ C1
L
LC
E
sin pt
þ C1 þ RD
E
ED
sin pt
sin pt ¼
RD
R D2
E
¼ 2 D sin pt
Rp
E
E
¼ 2 p cos pt ¼ cos pt:
Rp
Rp
¼
Ordinary Differential Equations
n
1.69
Thus, the complete solution is
Rt
E
ðc1 cos pt þ c2 sin ptÞ cos pt:
q ¼ 1
2L
Rp
the upward direction and with an applied external
force f (t) = 5 sin t. Find the subsequent motion of
the mass if the force due to air resistance is 90 x0 N.
Using the initial condition q = 0 for t = 0, we get
E
E
0 ¼ c1 or c1 ¼
:
Rp
Rp
Also
dq
Rt
¼ 1
ðc1 sin pt þ c2 cos ptÞp
i¼
dt
2L
R
E
ðc1 cos pt þ c2 sin ptÞþ sin pt:
2L
R
Using the initial condition i = 0 for t = 0, we get
Rc1
0 ¼ pc2 2L R E
¼ pc2 2L Rp
RE
¼ pc2 2Lp
E
and so c2 = 2Lp
2 . Hence, the solution is
Rt
E
2
i¼ 1
cos
pt
p
sin pt þ
Rp
2Lp2
2L
R E
E
E
cos pt þ
sin pt þ sin pt
2
2L Rp
2Lp
R
Et
¼
sin pt
2L
Solution. The describing differential equation is
d2x
a dx
k
1
dt2 þ m dt þ m x ¼ m f ðtÞ:
1.21
MASS-SPRING SYSTEM
In Example 1.4, we have seen that the differential
equation governing a Mass-spring system is
d2x
dx
m 2 þ a þ kx ¼ f ðtÞ
dt
dt
where m is the mass, a dx
dt the damping force due to
the medium, k is spring constant, and x represents
the displacement of the mass.
This is exactly the same differential equation
which occurs in LCR electric circuits. When a =
0, the motion is called undamped whereas if a 6¼ 0,
the motion is called damped. If f (t) = 0, then the
motion is called forced.
EXAMPLE 1.138
A mass of 10 kg is attached to a spring having spring
constant 140 N/m. The mass is started in motion from
the equilibrium position with a velocity of 1 m/sec in
Here a = 90, k = 140, and m = 10. Therefore,
we have
d2x
dx
1
þ 9 þ 14x ¼ sin t:
dt2
dt
2
The auxiliary equation is m2 + 9m + 14 = 0, and so
m = 2, 7. Therefore,
C:F: ¼ c1 e2t þ c2 e7t :
Further
1
1
1
sint ¼
sint
P:I: ¼ 2
D þ9Dþ14 2
2ð1þ9Dþ14Þ
1
1
1 9D13
sint ¼
sint
¼
2 ð9Dþ13Þ
2 ð81169Þ
1
¼
ð13sint þ9costÞ
500
13
9
¼
sint cost:
500
500
Hence, the complete solution is
13
9
sin t cos t:
x ¼ c1 e2t þ c2 e7t þ
500
500
Using the initial condition x(0) = 0, we get 0 = c1+c2
9
9
500
and so c1 + c2=500
. Since dx
dt (0) = 1 (initial
velocity in upper direction), we get
13
513
1 ¼ 2c1 7c2 þ
or2c1 þ7c2 ¼
:
500
500
90
99
, c2 = 500
. Hence
Solving for c1 and c2, we get c1 = 500
1
ð90e2t þ 99e7t þ 13 sin t 9 cos tÞ:
x ¼
500
EXAMPLE 1.139
If in a mass spring system, mass = 4kg, spring constant = 64, f (t) = 8 sin 4t, and if there is no air
resistance and initial velocity, then find the subsequent motion of the weight. Show that the resonance occurs in this case.
Solution. The governing equation is
d 2 x a dx k
1
þ
þ x ¼ f ðtÞ:
dt2 m dt m
m
1.70
n
Engineering Mathematics-II
Therefore, we have
d2x
þ16x¼ 2sin4t:
dt2
the mass of the particle, l be the length of the string,
and O be the fixed point of the string (Figure 1.9).
O
The auxiliary equation is m2 + 16 = 0 and so m =
± 4i. Therefore,
T
l
D
C:F: ¼ c1 cos4tþc2 sin4t:
B
P
s
A
Now
mg sin
1
2t
ð2 sin 4tÞ ¼
ðsin4tÞ
þ 16
2D
t Dðsin 4tÞ
t
¼
¼
ð4cos4tÞ
16
16
t
¼ cos4t :
4
P:I: ¼
¼ c1 cos 4tþc2 sin4t t
cos 4t:
4
Using initial condition x(0) ¼ 0, we have 0 ¼ c1.
Differentiating w.r.t. t, we get
dx
¼ 4c1 sin4tþ4c2 cos4t
dt
1
½cos4t 4tsin4t:
4
1
Now dx
dt ¼ 0 at t ¼ 0. Therefore, 0 ¼ 4c2 – 4 which
1
gives c2 ¼ 16. Hence
x
¼
1
t
sin 4t cos 4t:
16
4
We observe that x (t) ! 1 as t ! 1 due to the
presence of the term t cos 4t. This term is called a
secular term. The presence of secular term causes
resonance because the solution becomes unbounded.
1.22
mg cos
Figure 1.9
D2
Hence the complete solution is
x
mg
SIMPLE PENDULUM
The system in which a heavy particle (bob) is
attached to one end of a light inextensible string, the
other end of which is fixed, and oscillates under the
action of gravity force in a vertical plane is called a
simple pendulum. To describe its motion, let m be
Let P be the position of the heavy particle at any
time t and let ff AOP ¼ h, where OA is vertical
line through O. Then the force acting on the bob are
(a) weight mg acting vertically downward
(b) tension T in the string.
Resolving mg, we note that tension is balanced by mg
cosh. The equation of motion along the tangent is
m
d2s
¼ mg sin h
dt2
or
d2
ðlhÞ ¼ g sin h
dt2
or
d2h
g
¼ sin h
dt2
l
¼
g
h3
h þ :: (using expansion for sin hÞ
3!
l
¼
gh
to the first approximation:
l
Thus the differential equation describing the motion
of bob is
d2h
þ v2 h
dt2
¼ 0;
where v2 ¼ gl. The auxiliary equation of this equation is
m2 þ v2 ¼ 0
and so m ¼ ± vi. Therefore, the solution is
h ¼ c1 cos vt þ c2 sin vt
rffiffiffi
rffiffiffi
g
g
¼ c1 cos
t þ c2 sin
t:
l
l
Ordinary Differential Equations
n
The motion in case of simple pendulum is simple
harmonic motion where time period
sffiffiffi
2
l
¼ 2
T ¼
:
v
g
Also
The motion of the bob from one extreme position to
the other extreme position on the other side of A is
called a beat or a swing. Therefore,
qffiffi
Time for one swing ¼ 12 T ¼ gl .
But for t ¼ 0, dx
dt ¼ 0: Therefore, 0 ¼ nc2 F0
so c2 ¼ 2n
2.
Hence
F0
F0 t
cos nt
x ¼ 2 sin nt 2n
2n
F0
¼ 2 ðsin nt nt cos ntÞ:
2n
The pendulum which beats once every second is
called a second’s pendulum. Therefore, the number of beats in a second pendulum in one day is
equal to the number of seconds in a day. Thus, a
second pendulum beats 86,400 times a day.
Now, since, the time for 1 beat in a second’s
pendulum is 1 sec, we have
sffiffiffi
rffiffiffiffiffiffiffiffi
l
l
¼
1 ¼ g
981
and so l ¼ 99.4 cm is the length of second’s
pendulum.
EXAMPLE 1.140
The differential equation of a simple pendulum is
d2 x
2
dt2 þ v0 x ¼ F0 sin nt, where v0 and F0 are constants. If initially dx
dt ¼ 0, determine the motion
when v0 ¼ n
Solution. We have
2
d x
þ n2 x ¼ F0 sin nt since; v0 ¼ n:
dt2
The auxiliary equation is m2 + n2 ¼ 0 and so m ¼
± ni. Therefore,
C: F: ¼ c1 cos nt þ c2 sin nt:
Now
F0
F0 t
sin nt ¼
sin nt
2
þ
2D
Zn
F0 t
F0 t
sin nt dt ¼ cos nt:
¼
2
2n
P:I: ¼
D2
Thus, the complete solution is
x
¼ c1 cos ntþc2 sin nt F0 t
cos nt:
2n
Initially x ¼ 0 for t ¼ 0, so c1 ¼ 0.
1.23
1.71
dx
¼ nc1 sin ntþnc2 cos nt
dt
F0
½cos nt tn sin nt:
2n
F0
2n
and
SOLUTION IN SERIES
The method of series solution of differential equations is applied to obtain solutions of linear differential equations with variable coefficients.
Consider the differential equation
P0 ðxÞ
d2y
dy
þ P1 ðxÞ þ P2 ðxÞy ¼ 0;
2
dx
dx
ð117Þ
where P0 ðxÞ ; P1 ðxÞ, and P2 ðxÞ are polynomials in
x. This equation can be written as
d 2 y P1 ðxÞ dy P2 ðxÞ
y ¼ 0:
þ
þ
dx2 P0 ðxÞ dx P0 ðxÞ
ð118Þ
The point x ¼ a is called an ordinary point of the
P2 ðxÞ
equation (117) or (118) if the functions PP10 ðxÞ
ðxÞ and P0 ðxÞ
are analytic at x ¼ a. In other words, x ¼ a is an
ordinary point of (117) if P0 ðaÞ 6¼ 0.
P2 ðxÞ
If either (or both) of PP10 ðxÞ
ðxÞ or/and P0 ðxÞ is (are) not
analytic at x ¼ a, then x ¼ a is called a singular
point of the equation (117) or (118). Thus x ¼ a is a
singular point of (117) if P0(a) ¼ 0.
Further, let
P1 ðxÞ
;
Q1 ðxÞ ¼ ðx aÞ
P0 ðxÞ
P2 ðxÞ
:
Q2 ðxÞ ¼ ðx aÞ2
P0 ðxÞ
If Q1 and Q2 are both analytic at x ¼ a, then x ¼ a is
called a regular singular point of (117) otherwise it
is called irregular point of (117).
For example, consider the equation
d2y
dy
9xð1 xÞ 2 12 þ 4y ¼ 0:
dx
dx
1.72
n
Engineering Mathematics-II
We have
P1 ðxÞ
12
P2 ðxÞ
4
¼
¼
and
;
P0 ðxÞ 9xð1 xÞ
P0 ðxÞ 9xð1 xÞ
which are not analytic at x ¼ 0 and x ¼ 1. Hence
x ¼ 0 and x ¼ 1 are singular points of the given
equation. Further, at x ¼ 0,
12
4
; Q2 ðxÞ ¼
;
Q1 ðxÞ ¼
9ð1 xÞ
ð1 xÞ
which are analytic at x ¼ 0. Thus x ¼ 0 is a regular
singular point.
For x ¼ 1, we have
12
4
;
Q1 ðxÞ ¼ ; Q2 ðxÞ ¼
9x
9x
which are analytic if x ¼ 1. Hence x ¼ 1 is also
regular.
1.23.1 Solution About Ordinary Point
If x ¼ a is an ordinary point of the differential
equation (117), then its every solution can be
expressed in the form
y ¼ a0 þ a1 ðx aÞ þ a2 ðx aÞ2
ð119Þ
þ a3 ðx aÞ3 þ . . . ;
where the power series converges in some interval
|x – a| < R about a. Thus the series may be differentiated term by term on this interval and we have
dy
¼ a1 þ 2a2 ðx aÞ þ 3a3 ðx aÞ2 þ . . . ;
dx
d2y
¼ 2a2 þ 6a3 ðx aÞ þ . . . :
dx2
d2 y
Substituting the values of y ; dy
dx, and dx2 is (117), we
get an equation of the type
c0 þ c1 ðx aÞ þ c2 ðx aÞ2 þ . . . ¼ 0; ð120Þ
where the coefficients c0, c1, and c2 are functions of a.
Then (120) will be valid for all x in |x – a| < R if all
c1, c2,. . . are zero. Thus
c0 ¼ c1 ¼ c2 ¼ . . . :0:
ð121Þ
The coefficients ai of (119) are obtained from (121).
In case (120) is expressed in powers of x, then
equating to zero the coefficients of the various
powers of x will determine a2, a3, a4,. . . in terms of
a0 and a1. The relation obtained by equating to
zero the coefficient of xn is called the recurrence
relation.
1.23.2 Solution About Singular Point (Forbenious
Method)
If x ¼ a is a regular singular point of (117), then the
equation has at least one non-trivial solution of the
form
y ¼ ðxaÞm ½a0 þa1 ðxaÞþa2 ðxaÞ2 þ...; ð122Þ
where m is a definite constant (real or complex) and
the series on the right converges at every point of
the interval of convergence with centre a.
Differentiating (122) twice, we get
dy
¼ ma0 ðx aÞm1 þ ðm þ 1Þa1 ðx aÞm þ . . .
dx
d2y
¼ mðm 1Þa0 ðx aÞm2
dx2
þ mðm þ 1Þa1 ðx aÞm1 þ . . . :
Substituting the values of y;
get an equation of the form
dy
dx
2
and ddx2y in (117), we
c0 ðx aÞmþk þ c1 ðx aÞmþkþ1
þ c2 ðx aÞmþkþ2 þ . . . ¼ 0;
ð123Þ
where k is an integer and the coefficients ci are
functions of m and ai. In order that (123) be valid in
|x – a| < R, we must have
c0 ¼ c1 ¼ c2 ¼ . . . ¼ 0:
ð124Þ
On equating to zero the coefficient c0 in (123), we
get a quadratic equation in m, called indicial
equation, which gives the value of m. The two roots
m1 and m2 of the indicial equation are called
exponents of the differential equation (117). The
coefficients a1 ; a2 ; a3 ; . . . are obtained in terms of
a0 from c1 ¼ c2 ¼ :: ¼ 0. Putting the values of
a1 ; a2 ; . . .in (122), the solution of (117) is obtained.
If m1 m2 6¼ 0 or a positive integer, then the
complete solution of equation (117) is
y ¼ c1 ðyÞm1 þ c2 ðyÞm2 :
If m1 – m2 ¼ 0, that is, the roots of indicial equation
are equal, then the two independent solutions are
@y
.
obtained by substituting the value of m in y and @m
Thus, in this case,
@y
y ¼ c1 ðyÞm1 þ c2
@m m1
Ordinary Differential Equations
If m1 – m2 is a positive integers making a coefficient
of y infinite when m ¼ m2, then the form of y is
modified by replacing a0 by k(m – m2). Two independent solutions of the differential equation (117)
are then
@y
y ¼ c1 ðyÞm2 þ c2
@m m2
If m1 – m2 is a positive integer making a coefficient
of y indeterminate when m ¼ m2, then the complete
solution of (117) is
y ¼ c1 ðyÞm2 :
EXAMPLE 1.141
Find the power series solution of the equation
d2y
dy
ð1 x2 Þ 2 2x þ 2y ¼ 0
dx
dx
in powers of x, that is, about x ¼ 0.
Solution. Let
y ¼ a 0 þ a1 x þ a 2 x 2 þ a3 x 3 þ a4 x 4 þ . . . :
2
d y
Substituting the values of y, dy
dx and dx2 in the given
differential equation, we get
ð1 x2 Þ½2a2 þ 6a3 x þ . . . þ nðn 1Þan xn2 þ . . .
2x½a1 þ 2a2 x þ 3a3 x2 þ . . . þ nan xn1 þ . . .
þ 2ða0 þ a1 x þ a2 x2 þ . . . þ an xn þ . . .Þ ¼ 0
or
2ða2 þ a0 Þ þ ð6a3 2a1 Þx
þ ð12a4 2a2 4a2 þ 2a2 Þx
2
þ . . . þ ½ðn þ 2Þðn þ 1Þanþ2
nðn 1Þan 2nan þ 2an xn þ . . . ¼ 0
y ¼ a0 þa1 xþa2 x2 þa3 x3 þa4 x4 þa5 x5 þa6 x6 þ...
1
1
¼ a0 þa1 xa0 x2 a0 x4 a0 x6 þ...
3
5
2 1 4 1 6
¼ a0 1x x x ... þa1 x:
3
5
EXAMPLE 1.142
Find the solution in series of the equation
d2y
dy
þ x þ x2 y ¼ 0
2
dx
dx
about x ¼ 0.
Solution. The point x ¼ 0 is a regular point of the
given differential equation. So, let the required
solution be
y ¼ a0 þa1 xþa2 x2 þa3 x3 þa4 x4 þ...þan xn þ... :
Then differentiating twice, we get
dy
¼ a1 þ 2a2 x þ 3a3 x2 þ 4a4 x3 þ ... þ nan xn1 þ ...
dx
d2y
¼ 2a2 þ 6a3 x þ 12a4 x2 þ 20a5 x3
dx2
þ ... þ nðn 1Þan xn2 þ ... :
Substituting the values of y;
equation, we get
dy
dx,
2
and ddx2y in the given
½2a2 þ 6a3 x þ 12a4 x2 þ 20a5 x3
þ . . . þ nðn 1Þan xn2 þ . . .
þ x½a1 þ 2a2 x þ 3a3 x2 þ 4a4 x3
or
1.73
Equating to zero the coefficients of the various
powers of x, we get
1
a2 ¼ a0 ; a3 ¼ 0; a4 ¼ a0; ...
3
anþ2 ½n2 þ 3n þ 2 þ an ðn2 n þ 2Þ ¼ 0:
Taking n ¼ 3, 4, . . .
20a5 10a3 ¼ 0 ) a5 ¼ 0
1
30a6 18a4 ¼ 0 ) a6 ¼ 18
30 a4 ¼ 5 a0
...
...
Therefore,
Differentiating twice we get
dy
¼ a1 þ2a2 xþ3a3 x2 þ4a4 x3 þ...þnan xn1 þ...
dx
d2y
¼ 2a2 þ6a3 xþ...þnðn1Þan xn2 þ... :
dx2
n
2ða2 þ a0 Þ þ 6a3 x þ ð12a4 4a2 Þx2
þ . . . þ nan xn1 þ . . .
þ . . . þ ½ðn2 þ 3n þ 2Þanþ2 ðn2 nÞan
þ x2 ½a0 þ a1 x þ a2 x2 þ a3 x3
þ 2an ð1 nÞxn þ . . . ¼ 0:
þ a4 x4 þ . . . þ an xn þ . . . ¼ 0
1.74
n
Engineering Mathematics-II
or
2a2 þ ð6a3 þ a1 Þx þ ð12a4 þ 2a2 þ a0 Þx
2
þ ð20a5 þ 3a3 þ a1 Þx3 þ ð30a6 þ 4a4 þ a2 Þx4
þ ð42a7 þ 5a5 þ a3 Þx5 þ . . .
þ . . . ½ðn þ 2Þðn þ 1Þanþ2 þ nan þ an2 xn þ . . . ¼ 0:
Equating to zero the coefficients of the various
powers of x, we get
1
a2 ¼ 0; a3 ¼ a1 ;
6
Putting the values of y ;
equation, we get
dy
dx
and
d2y
dx2 in
the given
2a2 þ 6a3 x þ 12a4 x2 þ 20a5 x3 þ 30a6 x4
þ . . . þ nðn 1Þan xn2 þ . . .
þ x½a0 þ a1 x þ a2 x2 þ a3 x3 þ a4 x4 þ a5 x5 þ . . . ¼ 0
or
2a2 þ ð6a3 þ a0 Þx þ ð12a4 þ a1 Þx2 þ ð20a5 þ a2 Þx3
þ . . . þ ½ðn þ 2Þðn þ 1Þanþ2 þ anþ1 xn þ . . . :
Equating to zero the coefficients of various powers
1
of x, we get
12a4 þ2a2 þa0 ¼ 0; which yields a4 ¼ a0
12
1
a2 ¼ 0; 6a3 þ a0 ¼ 0 which yields a3 ¼ a0
1
6
20a5 þ3a3 þa1 ¼ 0 which yields a5 ¼ a1
40
1
12a4 þ a1 ¼ 0 which yields a4 ¼ a1
1
12
30a6 þ4a4 þa2 ¼ 0 which yields a6 ¼ a0 ; and so on:
90
20a5 þ a2 ¼ 0 which yields a5 ¼ 0
ðn þ 2Þðn þ 1Þanþ2 þ an1 ¼ 0:
Hence
1
1
Putting n ¼ 4, 5, 6, 7, . . ., we get
y ¼ a0 þ a1 x a1 x3 a0 x4
6
12
a3
1
1
1
a0
30a6 þ a3 ¼ 0 which yields a6 ¼ ¼
a1 x5 þ a0 x6 . . .
30
180
40
90
a
1
4
1
1
42a7 þ a4 ¼ 0 which yields a7 ¼ ¼
a1 ;
¼ a0 1 x 4 þ x 6 . . .
42
504
12
90
and so on:
1 3
1 5
þ a1 x x x . . . :
Hence
6
40
1 3
1 6
y ¼ a0 1 x þ
x ...
6
180
EXAMPLE 1.143
1 4
1 7
Find power series solution of the equation
x
x
x
þ
.
.
.
:
þ
a
1
12
504
d2y
þ xy ¼ 0
dx
EXAMPLE 1.144
in powers of x, that is, about x ¼ 0.
Find series solution of the differential equation
Solution. We note that x ¼ 0 is a regular point of the
d 2 y dy
x 2 þ y¼0
given equation. Therefore, its solution is of the form
dx
dx
y ¼ a 0 þ a1 x þ a 2 x 2 þ a3 x 3 þ a4 x 4 þ a 5 x 5 þ a6 x 6
about x ¼ 0.
þ . . . þ an x n þ :
Solution. The point x ¼ 0 is a regular singular point
Differentiating twice successively, we get
of the given equation. So, let
dy
y ¼ a0 xm þ a1 xmþ1 þ a2 xmþ2 þ . . . ; a0 6¼ 0:
¼ a1 þ 2a2 x þ 3a3 x2 þ 4a4 x3 þ 5a5 x4
dx
þ 6a6 x5 þ . . . þ nan xn1 þ . . .
Differentiating twice in succession, we get
2
d y
2
3
1
¼ 2a2 þ 6a3 x þ 12a4 x þ 20a5 x
dy X
dx2
¼
ðn þ mÞan xnþm1
dx n¼0
þ 30a6 x4 þ . . . þ nðn 1Þxn2 þ . . .
Ordinary Differential Equations
and
n
1.75
Therefore,
d2y
¼
dx2
1
X
ðn þ mÞðn þ m 1Þan xnþm2 :
n¼0
2
d y
Substituting the values of y ; dy
dx and dx2 in the given
equation, we get
1
X
ðn þ mÞðn þ m 1Þan xnþm2
x
n¼0
1
X
þ
a1 ¼
ðn þ mÞan x
nþm1
n¼0
1
X
a2 ¼
a3 ¼
1
ðm þ 1Þ2
1
ðm þ 2Þ
1
ðm þ 3Þ2
an x
¼0
n¼0
y ¼ a0 x m þ
þ
ðn þ mÞðn þ m 1Þan x
n¼0
þ
1
X
ðn þ mÞan xnþm1 n¼0
nþm1
1
X
an xnþm ¼ 0
n¼0
or
1
X
ðn þ mÞ2 an xnþm1 n¼0
1
X
an xnþm ¼ 0
n¼0
or
1
X
ðn þ m þ 1Þ2 anþ1 xnþm n¼1
1
X
an xnþm ¼ 0
n¼0
or
m2 a0 xm1 þ
1
X
ðn þ m þ 1Þ2 anþ1 xnþm
n¼0
1
X
an xnþm ¼ 0
n¼0
or
m2 a0 xm1 þ
1
X
xnþm
n¼0
½ðn þ m þ 1Þ2 anþ1 an ¼ 0
ð125Þ
Therefore, the indicial equation is m ¼ 0, which
yields m ¼ 0, 0. Equating to zero other coefficients
in (125), we get
2
ðn þ m þ 1Þ2 anþ1 ¼ an ;
a1 ¼
a2 ¼
1
2
ðm þ 1Þ ðm þ 2Þ2
1
a0 ;
ðm þ 1Þ2 ðm þ 2Þ2 ðm þ 3Þ2
a0 ;
and so on. Thus
nþm
1
ðmþ1Þ2
or
1
X
2
a0 ;
n
0:
a0 xmþ1 þ
1
ðmþ1Þ2 ðmþ2Þ2
a0 xmþ2
1
a0 xmþ3 þ...
ðmþ1Þ ðmþ2Þ2 ðmþ3Þ2
"
1
1
m
¼ a0 x 1þ
xþ
x2
ðmþ1Þ2
ðmþ1Þ2 ðmþ2Þ2
#
1
3
x :
þ
ðmþ1Þ2 ðmþ2Þ2 ðmþ3Þ2
2
Putting m ¼ 0, we get one solution of the given
differential equation as
1
1
y1 ¼ c1 1 þ x þ x2 þ x3 þ . . . :
4
36
Further,
"
@y
1
1
xþ
x2
¼ a0 xm logx 1 þ
2
2
@m
ðm þ 1Þ
ðm þ 1Þ ðm þ 2Þ2
#
1
3
þ
x þ ...
ðm þ 1Þ2 ðm þ 2Þ2 ðm þ 3Þ2
"
2x
1
m
þ a0 x 3
2
ðm þ 1Þ ðm þ 1Þ ðm þ 2Þ2
!
#
2
2
2
x þ ... :
þ
ðm þ 1Þ ðm þ 2Þ2
Therefore,
@y
1
1
¼ a0 log 1 þ x þ x2 þ x3 þ . . .
@m m¼0
4
36
1
þ a0 2x ð2 þ 1Þx2 þ . . . :
4
1.76
n
Engineering Mathematics-II
Hence the solution of the given equations is
@y
y ¼ c1 ðyÞm¼0 þ c2
@m m¼0
1
1
¼ ðc1 þ c2 log xÞ 1 þ x þ x2 þ x3 þ . . .
4
36
1
1
2c2 x þ
1 þ x2 þ . . . :
4
2
EXAMPLE 1.145
Find the series solution near x ¼ 0 of the differential
equation
d2y
dy
x2 2 þ x þ x2 y ¼ 0:
dx
dx
2
(This equation can also the written as x ddx2y þ dy
dx þ
xy ¼ 0 and is known as Bessel’s Equation of order
zero.)
Solution. The point x ¼ 0 is a regular singular point
of the given equation. So, let
1
X
an xnþm :
y¼
n¼0
Differentiating twice in succession, we get
1
dy X
¼
ðn þ mÞan xnþm1
dx n¼0
1
d2y X
¼
ðn þ mÞðn þ m 1Þan xnm2 :
dx2 n¼0
2
d y
Putting the values of y ; dy
dx and dx2 in the given
equation, we get
1
X
ðn þ mÞðn þ m 1Þan xnþm2
x
n¼0
1
X
þ
ðn þ mÞan xnþm1 þ x
n¼0
1
X
an xnþm ¼ 0
n¼0
or
1
X
ðn þ mÞðn þ m 1Þan xnþm1
n¼0
þ
1
X
ðn þ mÞan xnþm1 þ
n¼0
1
X
an xnþmþ1 ¼ 0
n¼0
or
1
X
n¼0
ðn þ mÞ2 an xnþm1 þ
1
X
n¼0
an xnþmþ1 ¼ 0
or
1
X
ðn þ m þ 1Þ2 anþ1 xnþm þ
n¼1
1
X
an xnþmþ1 ¼ 0
n¼0
or
m2 a0 xm1 þ
þ
1
X
1
X
ðn þ m þ 1Þ2 anþ1 xnþm
n¼0
an x
nþmþ1
¼ 0:
n¼0
Therefore, the indicial equation is m2 ¼ 0, which
yields m ¼ 0, 0. Equating to zero the coefficients of
powers of xm ; xmþ1 xmþ2 ; . . . ; we get
ðn þ m þ 1Þ2 a1 ¼ 0 which yields a1 ¼ 0;
a0
;
ðm þ 2Þ2 a2 þ a0 ¼ 0 and so a2 ¼ ðm þ 2Þ2
ðm þ 3Þ2 a3 þ a1 ¼ 0 and so a3 ¼ 0;
a2
ðm þ 4Þ2 a4 þ a2 ¼ 0 and so a4 ¼
ðm þ 4Þ2
a0
¼
;
ðm þ 2Þ2 ðm þ 4Þ2
and so on. Hence,
a0
a0
xmþ2 þ
xmþ4 ...
y ¼ a0 x m ðmþ2Þ2
ðmþ2Þ2 ðmþ4Þ2
"
#
1
1
m
2
4
¼ a0 x 1
x þ
x ... :
ðmþ2Þ2
ðmþ2Þ2
Further,
"
@y
1
m
¼ a0 x log x 1 x2
@m
ðm þ 2Þ2
þ
1
#
x ...
ðm þ 2Þ2 ðm þ 4Þ2
"
2x2
x4
þ a0 xm
ðm þ 2Þ3 ðm þ 2Þ2 ðm þ 4Þ2
2
2
þ
þ ... :
mþ2 mþ4
Therefore,
@y
1
1
¼ a0 log x 1 x2 þ x4 . . .
@m m¼0
4
64
þ a0
4
x2
3x4
þ ... :
4 2:4:16
Ordinary Differential Equations
n
1.77
Hence the solution of the given equation is
@y
y ¼ c1 ðyÞm¼0 þc2
@m m¼0
1 2
1
1
¼ c1 1 2 x þ 2 2 x4 2 2 2 x6 þ...
2
2 :4
2 :4 :6
1 2
1 4
1
þc2 logx 1 2 x þ 2 2 x 2 2 2 x6 þ...
2
2 :4
2 :4 :6
1 2
1
1 4
þa0 2 x 2 2 1þ x
2
2 :4
2
1
1 1 6
þ 2 2 2 1þ þ x ...
2 :4 :6
2 3
1
1
¼ ðc1 þc2 logxÞ 1 2 x2 þ 2 2 x4
2
2 :4
1
1 2
1
1 4
6
2 2 2 x þ... þa0 2 x 2 2 1þ x
2 :4 :6
2
2 :4
2
1
1 1
þ 2 2 2 1þ þ x6 þ... :
2 :4 :6
2 3
Solution. Evidently x ¼ 0 is a regular singular point
of the given equation. So, let
The solution
or
y¼
1
X
an xnþm ;
an 6¼ 0:
n¼0
Differentiating twice in succession, we get
1
dy X
ðn þ mÞan xnþm1 ;
¼
dx n¼0
1
d2y X
¼
ðn þ mÞðn þ m 1Þan xnþm2 :
dx2 n¼0
Substituting the values of y ;
equation, we get
2xð1 xÞ
1
X
dy
dx,
ðn þ mÞðn þ m 1Þan xnþm2
n¼0
1
X
þ ð1 xÞ
ðn þ mÞan xnþm1 þ 3
n¼0
1
1
1
c1 ðyÞm¼0 ¼c1 1 2 x2 þ 2 2 x4 2 2 2 x6 þ...
2
2 :4
2 :4 :6
2
1
X
2
and ddx2y in the given
1
X
an xnþm ¼ 0:
n¼0
ðn þ mÞðn þ m 1Þan xnþm1
n¼0
is called Bessel function of the first kind of order
zero and is represented by J0 ðxÞ, where as the
solution.
@y
1
1
¼ c2 log x 1 2 x2 þ 2 2 x4
c2
@m m¼0
2
2 :4
1
22 :42 :62
2
þ
2xð1 xÞ
d2y
dy
þ ð1 xÞ þ 3y ¼ 0:
2
dx
dx
1
X
ðn þ mÞan xnþm1 n¼0
1
X
þ3
1
X
ðn þ mÞan xnþm
n¼0
an xnþm ¼ 0
n¼0
or
1
X
ðn þ mÞð2n þ 2m 1Þan xnþm1
n¼0
is called Neumann function or Bessel function of
second kind of order zero and is denoted by Y0 ðxÞ.
EXAMPLE 1.146
Find series solution about x ¼ 0 of the differential
equations
ðn þ mÞðn þ m 1Þan xnþm
n¼0
x6 . . .
1 2
1
1 4
þ a0 2 x 2 2 1 þ x
2
2 :4
2
1
1 1
þ 2 2 2 1 þ þ x6 . . .
2 :4 :6
2 5
1
X
1
X
½ðn þ mÞð2n þ 2m 1Þ 3an xnþm ¼ 0
n¼0
or
1
X
ðn þ m þ 1Þð2n þ 2m þ 1Þanþ1 xnþm
n¼1
1
X
n¼0
½ðn þ mÞð2n þ 2m 1Þ 3an xnþm ¼ 0
1.78
n
Engineering Mathematics-II
For m ¼ 12, we have
or
2n2 þ 2m2 þ 4mn m n 3
an
ðn þ m þ 1Þð2n þ 2m þ 1Þ
2n2 þ 12 þ 2n 12 n 3
¼
an
n þ 32 ð2n þ 2Þ
mð2m 1Þa0 xm1
1
X
þ
ðn þ m þ 1Þð2n þ 2m þ 1Þanþ1 xnþm
anþ1 ¼
n¼0
1
X
½ðn þ mÞð2n þ 2m 1Þ 3an xnþm ¼ 0:
n¼0
Therefore, the indicial equation is
mð2m 1Þ ¼ 0; which yields m ¼ 0 ;
1
:
2
Equating to zero the other coefficients, we get
ðn þ m þ 1Þð2n þ 2m þ 1Þanþ1
¼ ½ðn þ mÞð2n þ 2m 1Þ 3an
¼ ½2n2 þ 2m2 þ 4nm m n 3an
2n2 þ n 3
an
ð2n þ 3Þðn þ 1Þ
ð2n þ 3Þðn 1Þ
n1
an ¼
an :
¼
ð2n þ 3Þðn þ 1Þ
nþ1
¼
Putting n ¼ 0, 1, 2, . . ., we get
a1 ¼ a0 ; a2 ¼ 0;
1
1
a3 ¼ a2 ¼ 0 ; a4 ¼ a3 ¼ 0;
4
2
and so on. Therefore, the solution corresponding to
m ¼ 12 is
1
3
5
1
3
1
7
y 2 ¼ a0 x 2 þ a1 x 2 þ a 2 x 2 þ a3 x 2 þ . . .
¼ a0 x2 a0 x2 ¼ a0 x2 ð1 xÞ:
or
anþ1 ¼
2n2 þ 2m2 þ 4nm m n 3
an :
ðn þ m þ 1Þð2n þ 2m þ 1Þ
For m ¼ 0, we get
anþ1
2n2 n 3
ðn þ 1Þð2n 3Þ
an ¼
an
¼
ðn þ 1Þð2n þ 1Þ
ðn þ 1Þð2n þ 1Þ
2n 3
¼
an ; n 0:
2n þ 1
Putting n ¼ 0, 1, 2, . . ., we have
1
a1 ¼ 3a0 ; a2 ¼ a1 ¼ a0 ;
3
1
1
3
3
a3 ¼ a2 ¼ a0 ; a4 ¼ a3 ¼ a0 ; and so on:
5
5
7
35
Therefore, the solution for m ¼ 0 is
y1 ¼ a0 x0 þ a1 x þ a2 x2 þ a3 x3 þ . . .
1
3
¼ a0 3a0 x þ a0 x2 þ a0 x3 þ a0 x4 þ . . .
5
35
1
3
¼ a0 1 3x þ x2 þ x3 þ x4 þ . . . :
5
35
Hence, the general solution of the given equation is
y ¼ c1 y1 þ c2 y2
1
3
¼ c1 1 3x þ x2 þ x3 þ x4 þ . . .
5
35
1
þ c2 x2 ð1 xÞ:
EXAMPLE 1.147
Find series solution about x ¼ 0 of the differential
equation
d2y
dy
5
x2 2 x x2 þ y ¼ 0:
dx
dx
4
Solution. The point x ¼ 0 is a regular singular point
of the given equation. So, let
1
X
an xnþm ; a0 6¼ 0:
y¼
n¼0
Differentiating twice in succession, we have
1
dy X
¼
ðn þ mÞan xnþm1 ;
dx n¼0
;
1
d2y X
nþm2
¼
ðn þ mÞðn þ m 1Þan x
:
dx
n¼0
Ordinary Differential Equations
n
1.79
2
d y
Putting the values of y; dy
Hence, for m ¼ 12, we have, from (127), a2 ¼ 12 a0 .
dx, and dx2 in the given differential equation, we have
Putting n ¼ 3 in (126), we have
1
1
X
X
1
5
5
ðn þ mÞðn þ m 1Þan xnþm ðn þ mÞan xnþm
3
3
a3 ¼ a1 or 0: a3 ¼ a1
n¼0
n¼0
2
2
4
1
1
X
X
5
xnþmþ2 an xnþm ¼ 0
and so a3 may be any constant. Further, from (127),
4
n¼0
n¼0
we have
or
1
a3
a0
;
a 4 ¼ a 0 ; a 5 ¼ ; a6 ¼
1
X
5
10
144
8
nþm
ðn þ mÞðn þ m 1Þ ðn þ mÞ an x
a3
4
; and so on:
a7 ¼
n¼0
280
1
X
nþmþ2
an x
¼0
Hence for m ¼ 12, the required solution is
n¼0
x2 x4
x6
1
or
...
y ¼ a0 x 2 1 2
8 144
1
X
5
5
ðn þ mÞðn þ m 2Þ an xnþm
x
x7
12
3
4
þ
a
þ
þ
.
.
.
:
ð128Þ
x
x
þ
3
n¼0
10 280
1
X
an2 xnþm ¼ 0
Since this solution contains two constants a0 and a3
n¼2
and a0 6¼ 0, this is general solution of the given
or
differential equation. By taking m ¼ 52, the solution
is
5
5
mðm2Þ a0 xm þ ðmþ1Þðm1Þ a1 xmþ1
4
4
x2
x4
5
2
þ
þ
.
.
.
:
1
þ
y
¼
x
1
X
5
10 280
ðnþmÞðnþm2Þ an an2 xnþm ¼ 0:
4
n¼2
Hence, (128) is general solution.
Therefore, the indicial equation is
EXAMPLE 1.148
5
5
1
Find series solution about x ¼ 0 of the differential
mðm 2Þ ¼ 0; which yields m ¼ ; :
4
2
2
equation
Equating to zero the other coefficients, we get
ðm þ 1Þðm 1Þ 5
a1 ¼ 0 and so a1 ¼ 0
4
and
ðn þ mÞðn þ m 2Þ 5
an ¼ an2 for n
4
For m ¼ 12, we get
1
5
5
n
n
an ¼ an2 ; n
2
2
4
2:
1
an2 ; n
nðn 3Þ
2; n 6¼ 3
d2y
dy
12 þ 4y ¼ 0:
dx
dx
Solution. The point x ¼ 0 is a regular singular point
of the given equation. Let
1
X
an xnþm ; a0 6¼ 0:
y¼
n¼0
2 ð126Þ
or
an ¼
9xð1 xÞ
ð127Þ
Differentiation of y with respect to x yields
1
dy X
ðn þ mÞan xnþm1 ;
¼
dx n¼0
1
d2y X
¼
ðn þ mÞðn þ m 1Þan xnþm2 :
dx2 n¼0
1.80
n
Engineering Mathematics-II
2
d y
Substituting the values of y ; dy
dx, and dx2 in the given
equation, we get
1
X
ðn þ mÞðn þ m 1Þan xnþm1
9
n¼0
9
1
X
ðn þ mÞðn þ m 1Þan xnþm
n¼0
1
X
12
ðn þ mÞan x
nþm1
þ4
n¼0
1
X
an x
nþm
¼0
ðn þ mÞð9n þ 9m 21Þan xnþm1
n¼0
1
X
1
2
2
a1 ¼ a0 ; a 2 ¼ a1 ¼ a0 ;
3
3
9
7
14
5
35
a0 ;
a3 ¼ a 2 ¼ a0 ; a 4 ¼ a 3 ¼
9
81
6
243
and so on. Thus, the solution corresponding to
m ¼ 0 is
x 2
14
35 4
x þ ... :
y1 ¼ a 0 1 þ þ x 2 þ x3 þ
3 9
81
243
n¼0
or
1
X
Putting n ¼ 0, 1, 2, 3,. . ., we obtain
½9ðn þ mÞðn þ m 1Þ 4an xnþm ¼ 0
For m ¼ 73, we have
9 n 73 n 73 1
3n 6
anþ1 ¼
an ; n
¼
3 n 73 þ 1 ð3n 7 4Þ 3n 4
0:
n¼0
Putting n ¼ 0, 1, 2, 3,. . ., we get
or
1
X
ðn þ m þ 1Þð9n þ 9m 12Þanþ1 x
n¼1
1
X
3
9
a1 ¼ a0 ; a2 ¼ 3a1 ¼ a0 ;
2
2
a3 ¼ 0; a4 ¼ 0; a5 ¼ 0; . . . :
nþm
½9ðn þ mÞðn þ m 1Þ 4an xnþm ¼ 0
n¼0
or
mð9m21Þa0 xm1 þ
1
X
Thus, the solution corresponding to m ¼ 73 is
3
9
7
y2 ¼ a0 x3 1 þ x þ x2 :
2
2
½ðnþmþ1Þð9nþ9m12Þanþ1 Hence, the general solution of the given differential
equation is
ð9ðnþmÞðnþm1Þ4Þan xnþm ¼ 0:
y ¼ c1 y 1 þ c 2 y 2
Therefore, the indicial equation is
x 2
14
35 4
¼ c1 1 þ þ x 2 þ x3 þ
x þ ...
7
3
9
81
243
mð9m 21Þ ¼ 0; which yields m ¼ 0; :
3
3
9 2
7
3
:
x
þ
x
x
1
þ
þ
c
2
We note that the roots are distinct and their differ2
2
ence is not an integer. Equating other coefficients
of the powers of x to zero, we get
EXAMPLE 1.149
n¼0
ðn þ m þ 1Þð9n þ 9m 12Þanþ1
¼ ½9ðn þ mÞðn þ m 1Þ 4an
Find series solution about x ¼ 0 of the differential
equation
or
anþ1 ¼
9ðn þ mÞðn þ m 1Þ 4
an ; n
ðn þ m þ 1Þð9n þ 9m 12Þ
3m þ 1
an ; n
3m þ 3
0:
d2y
dy
þ x y ¼ 0:
2
dx
dx
Solution. Since x ¼ 0 is a regular point of the given
equation, so let
For m ¼ 0, we get
anþ1 ¼
ð1 þ x2 Þ
0:
y ¼ a 0 þ a1 x þ a 2 x 2 þ . . . þ a n x n þ . . . :
Ordinary Differential Equations
Then,
dy
¼ a1 þ2a2 xþ3a3 x2 þ4a4 x3 þ...þnan xn1 þ...
dx
d2y
¼ 2a2 þ6a3 xþ12a4 x2 þ20a5 x3
dx2
þ...þnðn1Þan xn2 þ... :
2
d y
Substituting the values of y ; dy
dx, and dx2 in the given
equation, we get
ð1þx2 Þ½2a2 þ6a3 xþ12a4 x2 þ20a5 x3
þ...þnðn1Þan xn2 þx½a1 þ2a2 xþ3a3 x2 þ4a4 x3 þ...þnan xn1 þ...
½a0 þa1 xþa2 x2 þ...þan xn þ... ¼ 0:
Equating to zero the coefficients of various powers
of x, we get
a0
2a2 a0 ¼ 0 and so a2 ¼ ;
2
6a3 þ a1 a1 ¼ 0 and so a3 ¼ 0;
2a2 þ 12a4 þ 2a2 a2 ¼ 0 and so
1
1
a4 ¼ a 2 ¼ a 0 ;
4
8
6a3 þ 20a5 þ 3a3 þ a3 ¼ 0 and so a5 ¼ 0;
and in general,
nðn 1Þan þ ðn þ 2Þðn þ 1Þanþ2 þ nan nan ¼ 0
nðn þ 1Þ
an :
ðn þ 1Þðn þ 2Þ
Putting n ¼ 4, 5,. . ., we get
12
2
1
a6 ¼ a4 ¼ a4 ¼ a0
30
5
20
and so on. Thus, the required solution is
1 2 1 4
1 6
y ¼ a0 1 x x x . . . þ a1 x:
2
8
20
anþ2 ¼
1.81
solution of Bessel’s equation of order n. This
equation can be written as
d 2 y 1 dy
n2
¼ 0:
þ
1
þ
x2
dx2 x dx
2
Since 1x and 1 nx2 are not analytic at 0, it follows
that 0 is a singular point of the given equation.
2
But x 1x and x2 1 nx2 are analytic at 0.
Therefore, x ¼ 0 is a regular singular point of the
equation. So, let
1
X
am xmþr ; am 6¼ 0
y¼
m¼0
be the series solution of the equation about x ¼ 0.
Differentiating twice in succession, we get
1
dy X
ðm þ rÞam xmþr1
¼
dx m¼0
1
d2y X
¼
ðm þ rÞðm þ r 1Þam xmþr2 :
dx2 m¼0
2
d y
Putting the values of y; dy
dx, and dx2 in the given differential equation, we get
1
1
X
X
ðm þ rÞðm þ r 1Þam xmþr þ
ðm þ rÞam xmþr
m¼0
1
X
þ
or
n
m¼0
am xmþrþ2 n2
m¼0
or
1
X
1
X
am xmþr ¼ 0
m¼0
½ðm þ rÞ2 n2 am xmþr þ
m¼0
or
1
X
1
X
am xmþrþ2 ¼ 0
m¼0
½ðm þ rÞ2 n2 am xmþr þ
m¼0
1
X
am2 xmþr ¼ 0
m¼2
or
1.24
BESSEL’S EQUATION AND BESSEL’S FUNCTION
The equation
d2y
dy
x2 2 þ x þ ðx2 n2 Þy ¼ 0;
dx
dx
where n is a non-negative real number, is called
Bessel’s equation of order n. It occurs in problems
related to vibrations, electric fields, heat conduction, etc. For n ¼ 0, we have already found its
solution in Example 1.136. Now we find series
ðr2 n2 Þa0 xr þ ½ðr þ 1Þ2 n2 a1 xrþ1
1 X
þ
½ ðm þ rÞ2 n2 am þ am2 xmþr ¼ 0:
m¼2
Equating to zero the coefficient of lower power of x,
we get the indicial equation as
r2 n2 ¼ 0; which yields r ¼ n; n:
Equating other coefficients to zero, we get
½ðr þ 1Þ2 n2 a1 ¼ 0 or a1 ¼ 0
1.82
n
Engineering Mathematics-II
and
But y1 can be expressed is
2
½ðm þ rÞ n am þ am2 ¼ 0;
2
y1 ¼ a0 xn ðn þ 1Þ
which yields
am ¼
am2
ðm þ rÞ2 n2
; m
2:
þ
Putting m ¼ 2, 3, 4, 5, 6,. . ., we get
þ
a3 ¼ a5 ¼ a7 ¼ . . . ¼ 0
1
24 :2 !ðn þ 1Þðn þ 2Þ ðn þ 1Þ
1
a4 ¼ n
a6 ¼ n
¼ a0 xn ðn þ 1Þ
a0
ðr þ2Þ2 n2
ð1Þ2 a0
on
o
ðr þ2Þ2 n2 ðr þ4Þ2 n2
ðr þ2Þ2 n2
on
ð1Þ3 a0
ðr þ4Þ2 n2
þ
on
o
ðr þ6Þ2 n2
and so on. For r ¼ n, we have
a0
a0
;
¼
a1 ¼ 0; a2 ¼ 2
2
4ðn þ 1Þ
ðn þ 2Þ n
a0
;
a4 ¼ 2
4 2 !ðn þ 1Þðn þ 2Þ
a0
; and so on;
a6 ¼ 3
4 3 !ðn þ 1Þðn þ 2Þðn þ 3Þ
where as a2 ¼ a5 ¼ a7 ¼ . . . ¼ 0. Thus the solution
corresponding to r ¼ n is
x6 þ ...
1
1
x2
2
ðn þ 1Þ 2 ðn þ 2Þ
1
1
x4 þ 6
x6 þ ...
24 :2 !ðn þ 3Þ
2 :3 ! ðn þ 4Þ
¼ a0 xn ðn þ 1Þ
¼ a0 2n ðn þ 1Þ
¼
x4
26 :3 !ðn þ 1Þðn þ 2Þðn þ 3Þ ðn þ 1Þ
and
a2 ¼ 1
1
x2
ðn þ 1Þ 22 ðn þ 1Þ ðn þ 1Þ
1
X
ð1Þm x2m
22m :m! ðn þ m þ 1Þ
m¼0
x2mþn
ð1Þm
m! ðn þ m þ 1Þ 2
m¼0
1
X
x2mþn
ð1Þm
;
m! ðn þ m þ 1Þ 2
m¼0
1
X
ð131Þ
where a0 ¼ 2n 1ðnþ1Þ. The solution (131) is called
the Bessel’s function of the first kind of order n and
is denoted by Jn(x). Thus,
1
x2mþn
X
1
1
ð1Þm
x4
x2 þ 2
Jn ðxÞ ¼
: ð132Þ
4ðn þ 1Þ
4 :2 !ðn þ 1Þðn þ 2Þ
m! ðn þ m þ 1Þ 2
m¼0
1
x6 þ . . . : ð129Þ
3
4 :3 !ðn þ 1Þðn þ 2Þðn þ 3Þ
Replacing n by –n in Jn(x), we get
Similarly, for r ¼ –n, the solution is
1
x2mn
X
ð1Þm
1
1
Jn ðxÞ ¼
; ð133Þ
x4
x2 þ 2
y2 ¼ a0 xn 1 m ! ðn þ m þ 1Þ 2
4ð1 nÞ
4 :2 !ð1 nÞð2 nÞ
m¼0
1
x6 þ . . . : ð130Þ
3
4 :3 !ð1 nÞð2 nÞð3 nÞ
which is called Bessel’s function of the first kind of
We observe that y1 ¼ y2 for n ¼ 0. Further, y1 is order –n. Thus, the complete solution of the
meaningless if n is a negative integer and y is Bessel’s equation of order n may be expressed as
y 1 ¼ a0 x n 1 2
meaningless if n is a positive integer. Hence if n is
non-zero and non-integer, then the general solution
of the Bessel’s equation of order n is
y ¼ c1 y1 þ c2 y2 :
y ¼ c1 Jn ðxÞ þ c2 Jn ðxÞ;
whose n is not an integer.
ð134Þ
Ordinary Differential Equations
n
1.83
When n is an integer, let y ¼ u(x)Jn(x) be a solution EXAMPLE 1.150
of the Bessel’s equation of order n. Then
Show that
dy 0
Jn ðxÞ ¼ ð1Þn Jn ðxÞ:
0
¼ u ðxÞJn ðxÞþuðxÞJn ðxÞ
dx
Solution. We have
d 2 y 00
¼ u ðxÞJn ðxÞþJn0 ðxÞu0 ðxÞþu0 ðxÞJn0 ðxÞþuðxÞJn00 ðxÞ
1
x 2mþn
2
X
dx
ð1Þm
Jn ðxÞ ¼
00
0
0
00
¼ u ðxÞJn ðxÞþ2u ðxÞJn ðxÞþuÞxÞJn ðxÞ:
m! ðn þ m þ 1Þ 2
m¼0
2
d y
Putting the values of y ; dy
dx, and dx2 in the given
equation, we get
uðxÞ½x2 Jn00 ðxÞ þ xJn0 ðxÞ þ ðx2 n2 ÞJn ðxÞ
and
Jn ðxÞ ¼
þ x2 u00 ðxÞJn ðxÞ þ 2x2 u0 ðxÞJn0 ðxÞ þ xu0 ðxÞJn ðxÞ ¼ 0:
Since Jn ðxÞ is a solution of the given equation, we
have
x2 Jn00 ðxÞ þ xJn0 ðxÞ þ ðx2 n2 ÞJn ðxÞ ¼ 0:
Therefore, the above expression reduces to
x2 u00 ðxÞJn ðxÞ þ 2x2 u0 ðxÞJn0 ðxÞ þ xu0 ðxÞJn ðxÞ ¼ 0
1
X
m¼0
But for positive integer n,
ðn þ m þ 1Þ ¼ ðn þ mÞ! and
ðn þ m þ 1Þ ¼ ðm nÞ!
Therefore,
Jn ðxÞ ¼
or
u00 ðxÞ
J 0 ðxÞ 1
þ2 n þ ¼0
0
u ðxÞ
Jn ðxÞ x
B
and so uðxÞ ¼ B
u ðxÞ ¼ 2
xJn ðxÞ
0
Z
1
X
m¼0
ð1Þm x 2mþn
m!ðn þ mÞ! 2
and
or
d
d
d
ðlog u0 ðxÞÞ þ 2 ðlog Jn ðxÞÞ þ ðlog xÞ ¼ 0
dx
dx
dx
or
d
½logðxu0 ðxÞJn2 ðxÞÞ ¼ 0:
dx
Integrating this expression, we get
log xu0 ðxÞJn2 ðxÞ ¼ log B so that xu0 ðxÞJn2 ðxÞ ¼ B:
Thus
x2mn
ð1Þm
:
m! ðn þ m þ 1Þ 2
dx
þ A:
xJn2 ðxÞ
Jn ðxÞ ¼
1
X
m¼0
ð1Þm x2mn
:
m!ðm nÞ! 2
Since, (-n)! is infinite for n > 0, we have
Jn ðxÞ
¼
ð1Þm x2mn
m!ðm nÞ! 2
m¼0
1
X
1
X
ð1Þmþn x2mþn
; by changing m to m þ n
m!ðn þ mÞ! 2
m¼0
1
X
ð1Þm x2mþn
¼ ð1Þn Jn ðxÞ:
¼ ð1Þn
2
m!ðn
þ
mÞ!
m¼0
¼
Hence, the complete solution is
Z
dx
Jn ðxÞ
y ¼ uðxÞJn ðxÞ ¼ A þ B
EXAMPLE 1.151
2
xJ
n ðxÞ
Z
Show that
dx
d
¼
AJ
ðxÞ
þ
BY
ðxÞ;
¼ AJn ðxÞ þ BJn ðxÞ
n
n
(i) ½xn Jn ðxÞ ¼ xn Jn1 ðxÞ
xJn2 ðxÞ
dx
where
d
Z
(ii) ½xn Jn ðxÞ ¼ xn Jnþ1 ðxÞ
dx
dx
Yn ðxÞ ¼ Jn ðxÞ
x
xJn2 ðxÞ
(iii) Jn ðxÞ ¼ ½Jn1 ðxÞ þ Jnþ1 ðxÞ
2n
is called the Bessel function of the second kind of
1
order n or the Neumann function.
(iv) Jn0 ðxÞ ¼ ½Jn1 ðxÞ Jnþ1 ðxÞ.
2
1.84
n
Engineering Mathematics-II
These results are known as recurrence formulae for
the Bessel’s function Jn(x).
Solution. (i) We know that
1
x 2mþn
X
ð1Þm
:
Jn ðxÞ ¼
m! ðn þ m þ 1Þ 2
m¼0
Therefore,
xn Jn ðxÞ ¼
1
X
m¼0
ðmþnÞ
ð1Þm
x2
m! ðn þ m þ 1Þ 22mþn
and so
1
X
d n
ð1Þm
x2ðmþnÞ1
: 2mþn
ðx Jn ðxÞÞ ¼
m! ðn þ m þ 1Þ 2
dx
m¼0
¼ xn
xn1þ2m
ð1Þm
m! ðn 1 þ m þ 1Þ 2
m¼0
1
X
¼ xn Jn1 ðxÞ:
(ii) Multiplying the expression for J-n(x) by x-n
throughout and differentiating, we get (ii)
(iii) Part (i) implies
xn Jn0 ðxÞ þ nxn1 Jn ðxÞ ¼ xn Jn1 ðxÞ
and so
n
Jn0 ðxÞ þ Jn ðxÞ ¼ Jn1 ðxÞ:
x
Similarly, part (ii) yields
n
Jn0 ðxÞ þ Jn ðxÞ ¼ Jnþ1 ðxÞ:
x
Adding (135) and (136), we obtain
2n
Jn ðxÞ ¼ Jn1 ðxÞ þ Jnþ1 ðxÞ
x
or
x
Jn ðxÞ ¼ ½Jn1 ðxÞ þ Jnþ1 ðxÞ
2n
or
2n
Jnþ1 ðxÞ ¼ Jn ðxÞ Jn1 ðxÞ:
x
(iv) Subtracting (136) from (135) yields
2Jn0 ðxÞ ¼ Jn1 ðxÞ Jnþ1 ðxÞ
or
1
Jn0 ðxÞ ¼ ½Jn1 ðxÞ Jnþ1 ðxÞ:
2
ð135Þ
EXAMPLE 1.152
1
1
Show that e2xðz zÞ is the generating function of the
Bessel’s functions.
Solution. We have
e2xðz zÞ ¼ e2xz : e 2xz
"
2 2
1
1
z
x zþ
x
¼ 1þ
2!
2
2
#
1 r r
x z
þ ...
þ... þ 2
r!
"
1 1 1 2 z2
þ x
1þ x z
2!
2
2
1 r r #
2x z
þ... þ
:
r!
1
1
1
1
The coefficient of zn in this expansion is
1 n 1 nþ1 1 2 1 nþ2 1 2
x
2x
x
2x
2x
þ 2
þ 2
þ ...
n!
ðn þ 1Þ!
ðn þ 2Þ! 2!
2mþn
1
X
ð1Þm
1
x
¼
m!
ðm
þ
nÞ!
2
m¼0
¼
ð136Þ
1
1
X
m¼0
x 2mþn
ð1Þm
¼ Jn ðxÞ:
m! ðn þ m þ 1Þ 2
Thus
e2xðz zÞ ¼
1
1
1
X
zn Jn ðxÞ:
n¼1
Hence, e2xðz zÞ is the generating function of the
Bessel’s function.
1
1
EXAMPLE 1.153
Show that
rffiffiffiffiffi
2
sin x
(i) J12 ðxÞ ¼
x
rffiffiffiffiffi
2
cos x
(ii) J 12 ðxÞ ¼
x
R
1
(iii) xJ02 ðxÞdx ¼ x2 ½J02 ðxÞ þ J12 ðxÞ.
2
Ordinary Differential Equations
Solution. (i) Putting n ¼ 12 in the expression for Jn(x),
we have
n
1.85
Solution. We know that Jn(z) is a solution of the
Bessel’s equation
d2y
dy
ð137Þ
z2 2 þ z þ ðz2 n2 Þy ¼ 0:
dz
dz
x 2mþ12
ð1Þm
1
2
m¼0 m ! 2 þmþ1
Put z ¼ li x. Then
1
x12 X
ð1Þm x 2m
dy dy dx 1 dy
¼
¼ : ¼
2 m¼0 m ! mþ 32 2
dz dx dz li dx
"
#
x12 1
and
1 x 2
1 x 4
3 5
7
¼
þ
...
d2y
1 d2y
2 2 2 2
2! 2 2
¼ 2 2:
2
dz
li dx
"
x12
1
1 x 2
d2y
¼
Putting the values of z; dy
dz , and dz2 in (137), we get
2 12 12 32 12 12 2
!
2
#
1 dy
2 2 1 d y
x
x
þ ðl2i x2 n2 Þy ¼ 0:
þ
l
l
i
i
1
x 4
2 dx2
l
dx
l
i
þ 5 3 1 1
...
i
2: 2 : 2 : 2 2 2
or
x12 2
2
x2 x4
dy
2d y
1 1 þ ...
¼
þ x þ ðl2i x2 n2 Þy ¼ 0:
ð138Þ
x
3! 5!
2 2
dx2
dx
r
ffiffiffiffiffi
pffiffiffi
2
x3 x5
2
sinx:
Hence Jn(lix) is a solution of (138).
¼ pffiffiffipffiffiffi x þ ... ¼
x
3! 5!
x J12 ðxÞ ¼
1
X
(ii) Putting n ¼ 12 in the expression for J-n(x) and
proceeding as in part (i), we obtain part (ii).
(iii) We have
Z
xJ02 ðxÞdx
¼
¼
x2 2
J ðxÞ
2 0
Z
2
Z
2
Z
x 2
J ðxÞþ
2 0
x 2
J ðxÞþ
2 0
of Bessel’s functions is an orthogonal system, with
weight function x, on ½0; a; ae <.
x2
Solution. Let
2J0 ðxÞJ00 ðxÞdx (integration by parts)
2
ui ¼ Jn ðli xÞ and uj ¼ Jn ðlj xÞ;
x2 J0 ðxÞJ1 ðxÞdx since J00 ðxÞ ¼ J1 ðxÞ
d
ðxJ1 ðxÞÞdx since
dx
d
ðxJ1 ðxÞÞ ¼ xJ0 ðxÞ
dx
x2
1
x2
¼ J02 ðxÞþ ðxJ1 ðxÞÞ2 ¼ ½J02 ðxÞþJ12 ðxÞ:
2
2
2
¼
xJ1 ðxÞ:
EXAMPLE 1.154
Show that Jn(lix) is a solution of
2
EXAMPLE 1.155
If l1, l2,. . . are the roots of Jn ðaxÞ ¼ 0 ; a e <,
show that the system
Jn ðl1 xÞ; Jn ðl2 xÞ; Jn ðl3 xÞ; . . .
d y
dy
þ x þ ðl2i x2 n2 Þy ¼ 0:
x2
dx
dx
i 6¼ j:
Then, by Example 1.154, we have
x2 u00i þ xu0i þ ðl2i x2 n2 Þui ¼ 0
ð139Þ
and
0
x2 u00j þ xu þ ðl2j x2 n2 Þuj ¼ 0:
J
ð140Þ
Multiplying (139) by uJ and (140) by ui and subtracting, we get
x2 uj u00i ui u00j þx uj u0i ui u0j þx2 ui uj l2i l2j ¼ 0
or
x uj u00i ui u00j þ uj u0i ui u0j ¼ x l2j l2i ui uj
1.86
n
Engineering Mathematics-II
or
0
d 0
0
0
x
uj ui ui uj þ uj ui ui uj ¼ x l2j l2i ui uj
dx
or
Or, using L’Hospital’s Rule,
Za
a2 li J 0 n ðlj aÞJ 0 n ðli aÞ
xJn2 ðli xÞdx ¼ lim
lj !li
2lj
0
i
d h 0
0
x uj ui ui uj ¼ x l2j l2i ui uj :
dx
a2 0
2
½J n ðli aÞ
2
a2 2
¼ Jnþ1
ðli aÞ:
2
¼
ð141Þ
Integrating (141) with respect to x in [0, a], we get
h 0
ia Za
0
2
2
x uj ui ui uj
¼ lj l i
xui uj dx
Hence
0
or
2 ðl aÞ
a2 Jnþ1
i
h
0
i
0
a li Jn lj a Jn ðli aÞ lj Jn ðli aÞJn lj a
¼
l2j
l2i
Za
ð142Þ
xui uj dx:
0
Since Jn lj a ¼ Jn ðli aÞ ¼ 0, it follows that
Za
xui uj ¼ 0; li 6¼ lj
l2j l2i
Za
xJn ðli xÞJn lj x dx ¼ 0;
xJn2 ðli xÞdx ¼ 1:
0
EXAMPLE 1.156
Show that xnJn(x) is a solution of the differential
equation
d2y
dy
x 2 þ ð1 2nÞ þ xy ¼ 0:
dx
dx
Solution. Let y ¼ xn Jn ðxÞ. Then
dy
d
¼ ðxn Jn ðxÞÞ ¼ xn Jn1 ; and
dx dx
0
or
Za
2
0
d2y
d
0
¼ ðxn Jn1 Þ ¼ xn Jn1 þ nxn1 Jn1 :
dx2 dx
i 6¼ j:
0
Hence the system {Jn(lix)} forms an orthogonal
system with weight x.
Putting the values of y;
equation, we get
dy
dx,
and
d2 y
dx2
in the given
d2y
dy
þ ð1 2nÞ þ xy
dx2
dx
n1
0
Jn1 þ xnþ1 Jn
¼ xnþ1 Jn1
x
x
Remark 1.3. From (142), we have
Za
xJn ðli xÞJn ðlj xÞdx
¼ xnþ1 ðJn Þ þ xnþ1 Jn ¼ 0:
0
¼
a½li Jn ðlj aÞJ 0 n ðli aÞ lj Jn ðli aÞJ 0 n ðlj aÞ
:
l2j l2i
If li ¼ lj , then right hand side of the above
expression is of 00 form. We assume that li is a
root of Jn ðaxÞ ¼ 0 and lj ! li . Then we have
Z a
xJn ðli xÞJn ðlj xÞdx
lim
lj !li
0
ali Jn ðlj aÞJ 0 n ðli aÞ
¼ lim
lj!li
l2j l2i
EXAMPLE 1.157
Express J4(x) is term of J0(x) and J1(x).
Solution. We know that
x
Jn ðxÞ¼ ½Jn1 ;ðxÞþJnþ1 ðxÞ ðrecurrence formulaÞ
2n
or
Jnþ1 ðxÞ ¼
2n
Jn ðxÞ Jn1 ðxÞ:
x
Ordinary Differential Equations
Putting n ¼ 1, 2, and 3, we get
2
J2 ðxÞ ¼ J1 ðxÞ J0 ðxÞ
x
4
J3 ðxÞ ¼ J2 ðxÞ J1 ðxÞ
x
4 2
¼
J1 ðxÞ J0 ðxÞ J1 ðxÞ
x x
8
4
¼ 2 J1 ðxÞ J0 ðxÞ J1 ðxÞ
x
x
8
4
¼ 2 1 J1 ðxÞ J0 ðxÞ;
x
x
6
J4 ðxÞ ¼ J3 ðxÞ J2 ðxÞ
x 6 8
4
¼
1 J1 ðxÞ J0 ðxÞ
x x
x
2
J1 ðxÞ J0 ðxÞ
x
48
6
24
2
¼ 3 J1 ðxÞ J1 ðxÞ 2 J0 ðxÞ J1 ðxÞ þ J0 ðxÞ
x
x
x
x
48 8
24
¼ 3 J1 ðxÞ þ 1 2 J0 ðxÞ:
x x
x
EXAMPLE 1.158
Solve
d2z
dz
x2 2 þ x þ ðx2 64Þz ¼ 0:
dx
dx
Solution. The given equation is evidently Bessel’s
equation of order 8, which is an integer. Therefore,
its general solution is
z ¼ c1 J8 ðxÞ þ c2 Y8 ðxÞ;
where c1 and c2 are arbitrary constants.
EXAMPLE 1.159
Solve the differential equation
d 2 y dy 1
x 2 þ þ y ¼ 0:
dx
dx 4
pffiffiffi
Solution. Let z ¼ x. Then
dy dy dz dy 1
¼ : ¼ : pffiffiffi ; and
dx dz dx dz 2 x
2
d 2 y dy 1
1 3
1
d y 1
2
p
ffiffi
ffi
p
ffiffi
ffi
þ
¼
:
:
x
dx2 dz 2
2
2 x dz2 2 x
1 d2y
1 dy
1 d2y
1 dy
¼
3 ¼ 2 2 3 :
2
4x dz
4z dz
4x2 dz 4z dz
n
1.87
Putting these values of the partial derivatives in the
given equation, we get
z2
1 d2y
1 dy
1 dy 1
þ
þ y¼0
4z2 dz2 4z3 dz
2z dz 4
or
d 2 y 1 dy 2 dy
þ
þy¼0
dz2 z dz z dz
or
z
or
z2
d 2 y dy
þ þ zy ¼ 0
dz2 dz
d2y
dy
þ z þ ðz2 02 Þy ¼ 0;
dz2
dz
which is Bessel’s equation of order 0. Therefore, its
general solution is
pffiffiffi
pffiffiffi
y ¼ c1 J0 ðzÞ þ c2 y0 ðzÞ ¼ c1 J0 x þ c2 Y0 x :
EXAMPLE 1.160
Show that
x
1
1
1
e2ðz zÞ ¼ J0 ðxÞ þ z J1 ðxÞ þ z2 þ 2 J2 ðxÞ
z
z
1
þ z3 3 J3 ðxÞ þ . . .
z
Hence or otherwise show that
Z
1
cosðx cos hÞdh:
J0 ðxÞ ¼
0
Solution. We know that
1
X
x
1
zn Jn ðxÞ ðgenerating functionÞ
e2ðz zÞ ¼
n1
¼ J0 þ
1
X
ðJn zn þ Jn zn Þ
n¼1
¼ J0 þ
1
X
ðJn zn þ ð1Þn Jn zn Þ
n¼1
¼ J0 þ
1
X
½zn þ ð1Þn zn Jn
n¼1
1
¼ J0 þ z J1 þ ðz2 þ z2 ÞJ2
z
þ ðz3 z3 ÞJ3 þ . . .
ð143Þ
1.88
n
Engineering Mathematics-II
Putting z ¼ ei we get z 1z ¼ 2i sin ,
z2 þ z12 ¼ 2 cos 2, and so on. Therefore, (143)
reduces to
eiðx sin Þ ¼ J0 þ ð2i sin ÞJ1 þ ð2 cos 2ÞJ2 þ . . .
or
cosðx sin Þ þ i sinðx sin Þ
¼ J0 þ ð2i sin ÞJ1 þ ð2 cos 2ÞJ2 þ . . . :
Separating real and imaginary parts, we get
cosðx sin Þ ¼ J0 þ ð2 cos 2ÞJ2
þ ð2 cos 4ÞJ4 þ . . .
ð144Þ
Thus, the given equation reduces to
z2
which is Bessel’s equation of order zero. Its solution is
3 pffiffiffiffiffiffi
y ¼ J0 ðzÞ ¼ J0 ð i xÞ ¼ J0 i2 x
¼1
þi
ð145Þ
which are called Jacobi’s series.
Putting ¼ 2 h in (144), we get
cosðx cos hÞ ¼ J0 ð2 cos 2hÞJ2 þ . . . :
i3 x2
i 6 x4
i 9 x6
i12 x8
þ
þ
...
22
24 ð2!Þ2 26 ð3!Þ2 28 ð4!Þ2
¼ 1
and
sinðx sin Þ ¼ ð2 sin ÞJ1 þ ð2 sin 3ÞJ3
þ ð2 sin 5ÞJ5 þ . . . ;
d2y
dy
þ z þ ðz2 0Þy ¼ 0;
dz2
dx
x4
22 :42
where
0
0
J0 ¼
1
ð1Þm
x4m
22 :42 :62 . . . ð4mÞ2
is called Bessel’s real (or ber) function and
ð2 cos 2hÞJ2 dh þ . . . ¼ J0 :
Hence
1
X
m¼1
Z
J0 dh ...
¼ ber x þ i bei x;
0
¼
x8
22 :42 :62 :82
x2
x6
x10
2 2 2 þ 2 2 2 2 2 ...
2
2
2 :4 :6
2 :4 :6 :8 :10
ber x ¼ 1 þ
Therefore,
Z
cosðx cos hÞdh
Z
þ
bei x ¼ 1
X
ð1Þm
m¼1
x4m2
22 :42 :62 . . . ð4m 2Þ2
called Bessel’s imaginary (or bei) function.
Z
cosðx cos hÞdh:
0
EXAMPLE 1.161
Solve the differential equation
d 2 y dy
x 2 þ ixy ¼ 0:
dx
dx
Solution. The given equation can be written as
d2y
dy
x2 2 þ x ix2 y ¼ 0:
dx
dx
pffiffiffiffiffiffi
Putting z ¼ i x, we have
dy dy dz pffiffiffiffiffiffi dy
¼ : ¼ i
;
dx dz dx
dz
d2y
d2y
¼
i
:
dx2
dz2
1.25
FOURIER–BESSEL EXPANSION
OF A CONTINUOUS FUNCTION
We have proved in Example 1.155 that Bessel’s
functions Jn ðlxÞ form an orthogonal set with
weight x. This property allows us to expand a
continuous function f in Fourier–Bessel series in a
range 0 to a. So, let
f ðxÞ ¼ c1 Jn ðl1 xÞ þ c2 Jn ðl2 xÞ þ þ cn Jn ðln xÞ
ð146Þ
where li are the roots of the equation Jn ðlaÞ ¼ 0.
To determine the coefficients ci , we multiply both
sides of the equation (146) by xJn ðli xÞ and integrate
with respect to x between the limits 0 to a. Using
orthogonal property and the remark cited above,
we have
Ordinary Differential Equations
Za
0
xJn ðli xÞdx
1¼
0
¼ ci
a2 2
J ðli aÞ :
2 nþ1
Hence
ci ¼
Za
2
2 ðl aÞ
a2 Jnþ1
i
1
X
cn J0 ðln xÞ:
1
Since f ðxÞ ¼ 1 and a ¼ 1, the coefficients ci are
given by
Z1
2
ci ¼ 2
xJ0 ðli xÞdx
J1 ðli Þ
0
xf ðxÞJn ðli xÞdx:
2
xJ1 ðli xÞ
¼ 2
li
J1 ðli Þ
2
:
¼
li J1 ðli Þ
0
Substituting the values of ci in the expression (146),
we get the Fourier–Bessel series of f.
1
0
EXAMPLE 1.162
Expand f ðxÞ ¼ x2 ð0 < x < aÞ in Fourier–Bessel
series in J2 ðln xÞ if ln a are the positive roots of
J2 ðxÞ ¼ 0.
Hence the required Fourier–Bessel series is
Solution. Let the Fourier–Bessel series of the function f ðxÞ ¼ x2 ð0 < x < aÞ be
1
X
cn J2 ðln xÞ:
x2 ¼
1.26
1¼2
The differential equation
ð1 x2 Þ
We know that
ci ¼
¼
Za
2 ðl aÞ
a2 Jnþ1
i
2
a2 J32 ðli aÞ
xf ðxÞJn ðli xÞdx
0
Za
x3 J2 ðli xÞdx
0
2
x3 J3 ðli xÞ
¼ 2 2
li
a J3 ðli aÞ
2a
:
¼
li J3 ðli aÞ
a
0
Hence the Fourier–Bessel series for f ðxÞ ¼ x2 is
1
X
J0 ðln xÞ
:
l J ðl Þ
n¼1 n 1 n
LEGENDRE’S EQUATION AND LEGENDRE’S
POLYNOMIAL
n¼1
2
1.89
Solution. Let the Fourier–Bessel series of f ðxÞ ¼ 1 be
Za
xf ðxÞJn ðli xÞdx ¼ ci
n
d2y
dy
2x þ nðn þ 1Þy ¼ 0
dx2
dx
ð147Þ
is called Legendre’s equation of order n, where n is
a real number. The equation can be written is the
form
d2y
2x dy nðn þ 1Þ
y ¼ 0:
þ
dx2 1 x2 dx
1 x2
nðnþ1Þ
2x
Since 1x
2 and 1x2 are analytic at 0, the point
x ¼ 0 is a regular point of the Legendre’s equation.
So let
1
X
am x m
y¼
m¼0
1
X
J2 ðln xÞ
:
x ¼ 2a
J ðl aÞ
l
n¼1 n 3 n
be the power series solution of the given Legendre’s
equation about x ¼ 0. Then
EXAMPLE 1.163
Expand f ðxÞ ¼ 1 in Fourier–Bessel series in J0 ðln xÞ
if l1 ; l2 ; :::; ln are the roots of J0 ðxÞ ¼ 0.
1
d2y X
¼
mðm 1Þam xm2 :
2
dx
m¼2
2
1
dy X
¼
mam xm1 ; and
dx m¼1
1.90
n
Engineering Mathematics-II
Putting the values of y; dy
dx, and
equation, we get
1
X
mðm 1Þam xm2
ð1 x2 Þ
d2y
dx2
in the given
2x
mam xm1 þ nðn þ 1Þ
m¼1
or
1
X
mðm 1Þam x
m2
m¼2
2
1
X
mam xm þ nðn þ 1Þ
mðm 1Þam x
m
1
X
am x m ¼ 0
m¼0
ðm þ 2Þðm þ 1Þamþ2 xm m¼0
2
am x m ¼ 0
m¼2
m1
or
1
X
1
X
m¼0
1
X
1
X
mðm 1Þam xn
m2
1
X
mam xm þ nðn þ 1Þ
m¼1
1
X
am xm ¼ 0
m¼0
or
½2a2 þ nðn þ 1Þa0 þ ½6a3 2a1 þ nðn þ 1Þa1 x
1
X
þ
½ðm þ 2Þðm þ 1Þamþ2
m¼2
þ ðn mÞðn þ m þ 1Þam xm ¼ 0:
Equating to zero the coefficient of powers of x, we
get
2a2 þ nðn þ 1Þa0 ¼ 0; which yields
nðn þ 1Þ
a0 ;
a2 ¼ 2!
6a3 2a1 þ nðn þ 1Þa1 ¼ 0; which gives
ðn 1Þðn þ 2Þ
a1 ;
a3 ¼ 3!
and in general
amþ2 ¼ ðn mÞðn þ m þ 1Þ
am ;
ðm þ 1Þðm þ 2Þ
m
2:
Putting m ¼ 0, 1, we observe that a2 and a3 are the
same as found above. Therefore,
amþ2 ¼ ðn 2Þnðn þ 1Þðn þ 3Þ
a0 ;
4!
ðn 3Þðn 1Þðn þ 2Þðn þ 4Þ
a5 ¼
a1
5!
and so on. Hence
a4 ¼
m¼2
1
X
which is called the recurrence solution of the
Legendre’s equation. Putting m ¼ 2, 3, . . . we have
ðn mÞðn þ m þ 1Þ
am ;
ðm þ 1Þðm þ 2Þ
m
0;
nðnþ1Þ 2 ðn1Þðnþ2Þ 3
a0 x x
2!
3!
ðn2Þnðnþ1Þðnþ3Þ 4
a0 x
þ
4!
ðn3Þðn1Þðnþ2Þðnþ4Þ 5
a1 x þ...
þ
5!
nðnþ1Þ 2 ðn2Þnðnþ1Þðnþ3Þ 4
¼a0 1
x þ
x þ...
2!
4!
ðn1Þðnþ2Þ 3
þa1 x
x
3!
ðn3Þðn1Þðnþ2Þðnþ4Þ 5
þ
x þ...
4!
¼a0 y1 ðxÞþa1 y2 ðxÞ;say:
y¼a0 þa1 x
Thus, y1 contains only even powers of x while y2
contains odd powers of x. Also y1 and y2 are linearly
independent.
If n is even, then y1(x) is a polynomial of degree
n and if n is odd, then y2(x) is a polynomial of
degree n. Moreover if an ¼ ð2nÞ!
2 , then
n
2 ðn!Þ
Pn ðxÞ ¼
a0 þ a2 x2 þ . . . þ an xn if n is even
a1 x þ a3 x3 þ . . . þ an xn if n is odd
is called the Legendre polynomial of degree n.
We note that
P0 ðxÞ ¼ 1; P1 ðxÞ ¼ x;
1
1
P2 ðxÞ ¼ ð3x2 1Þ; P3 ðxÞ ¼ ð5x2 3xÞ;
2
2
1
4
2
P4 ðxÞ ¼ ð35x 30x þ 3Þ; and so on:
8
Evidently, the value of the polynomial Pn(x), n ¼ 0,
1, 2,. . . is 1 for x ¼ 1.
EXAMPLE 1.164
Show that
1 dn 2
ðx 1Þn ðRodrigue0 s formulaÞ:
Pn ðxÞ ¼
n! 2n dxn
Ordinary Differential Equations
Solution. Putting u ¼ (x2 – 1)n, we have
du
2nxu
¼ 2nxðx2 1Þn1 ¼ 2
dx
x 1
and so
du
ð148Þ
ð1 x2 Þ þ 2nxu ¼ 0:
dx
Differentiating (148), (n + 1) times by Leibnitz’s
theorem, we get
þ 2n½xunþ1 þ ðn þ 1Þun ¼ 0
ð1 x2 Þunþ2 2xunþ1 þ nðn þ 1Þun ¼ 0
or
d 2 un
dun
þ nðn þ 1Þun ¼ 0:
2x
2
dx
dx
n
dn
2
satisfies the Legendre’s
Thus, un ¼ dx
n ðx 1Þ
equation
ð1 x2 Þ
d2y
dy
2x þ nðn þ 1Þy ¼ 0:
2
dx
dx
Since the solution of Legendre’s equation is Pn(x),
we have
dn
Pn ðxÞ ¼ cun ¼ c n ðx2 1Þn :
ð149Þ
dx
Putting x ¼ 1 in (149), we get
dn
1 ¼ c n ðx 1Þn ðx þ 1Þn
dx
x¼1
ð1 x2 Þ
n
¼ c½n! ðx þ 1Þ þ term containing x 1
1.91
Solution. We have, by binomial expansion,
1
½1ð2xt t2 Þ2
1
1:3
¼ 1þ ð2xt t2 Þþ 2 ð2xtt2 Þ2 þ...
2
2 :2!
1:3...ð2n5Þ
ð2xtt2 Þn2
þ n2
2 ðn2Þ!
þ
ð1 x2 Þunþ2 2ðn þ 1Þxunþ1 nðn þ 1Þun
or
n
1:3...ð2n3Þ
ð2xtt2 Þn1
2n1 ðn1Þ!
1:3...ð2n1Þ
ð2xt t2 Þn þ...
þ
2n :n !
1 3
¼ 1þxt þ þ :4x2 t2 þ...
2 8
þ
þ
1:3...ð2n1Þ
1:3...ð2n3Þ
ð2xÞn n1
ðn1Þð2xÞn2
n
2 :n!
2 ðn1Þ!
1:3...ð2n5Þ ðn2Þðn3Þ
ð2xÞn4 þ... tn þ...
2n2 ðn2Þ!
1:2
1
¼ 1þxt þ ð3x2 1Þt2 þ...
2
1:3...ð2n1Þ n nðn1Þ n2
þ
x x
n!
2ð2n1Þ
þ
nðn1Þðn2Þðn3Þ n4
x þ... tn þ...
2:4ð2n1Þð2n3Þ
¼ P0 ðxÞþtP1 ðxÞþt2 P2 ðxÞþ...þtn Pn ðxÞþ...
¼
1
X
tn Pn ðxÞ:
n¼0
Hence, ð1 2xt þ t2 Þ 2 generates Legendre polynomials Pn(x).
1
along with its powers]x¼1
¼ c n! 2n ;
and so c ¼ n!12n . Thus,
Pn ðxÞ ¼
1 dn 2
ðx 1Þn :
n! 2n dxn
EXAMPLE 1.166
Express the polynomial x3 + 2x2 – x – 3 in terms of
Legendre’s polynomials.
Solution. We know that
EXAMPLE 1.165
Show that
P0 ðxÞ ¼ 1
ð1 2xt þ t2 Þ2 ¼
1
1
X
1
tn Pn ðxÞ; j x j < 1; j t j < :
3
n¼0
[Thus ð1 2xt þ t2 Þ 2 is a generating function of
the Legendre’s polynomials.]
1
;
P1 ðxÞ ¼ x;
1
2
1
P2 ðxÞ ¼ ð3x2 1Þ and so x2 ¼ P2 ðxÞ þ ;
2
3
3
1
2
3
P3 ðxÞ ¼ ð5x3 3xÞ and so x3 ¼ P3 ðxÞ þ P1 ðxÞ:
2
5
5
1.92
n
Engineering Mathematics-II
or
Hence,
x þ 2x x 3
3
¼
2
Z1
Pm ðxÞPn ðxÞdx ¼ 0;
2
3
2
1
P3 ðxÞ þ P1 ðxÞ þ 2 P2 ðxÞ þ
5
5
3
3
P1 ðxÞ 3P0 ðxÞ
m 6¼ n:
1
Hence, {Pn(x)} forms an orthogonal system.
2
4
2
7
¼ P3 ðxÞ þ P2 ðxÞ P1 ðxÞ P0 ðxÞ:
5
3
5
3
EXAMPLE 1.167
Show that Legendre polynomials are orthogonal on
the interval [– 1, 1].
Solution. Let Pm and Pn be Legendre polynomials for
m 6¼ n. We want to show that
Z1
Pm ðxÞPn ðxÞdx ¼ 0; m 6¼ n:
1
EXAMPLE 1.168
nqffiffiffiffiffiffiffiffi
o
2nþ1
ðxÞ
forms an orthonormal
P
Show that
n
2
system.
nqffiffiffiffiffiffiffiffi
o
Solution. The system of polynomials 2nþ1
2 Pn ðxÞ will
form an orthonormal system if
2n þ 1
2
Z1
0 for m 6¼ n
1 for m ¼ n:
Pm ðxÞPn ðxÞdx ¼
1
By Example 1.167, it follows that
Since Pm and Pn satisfy Legendre’s equation, we
have
ð1x2 ÞP00n ðxÞ2xPm0 ðxÞþmðmþ1ÞPm ðxÞ ¼ 0 ð150Þ
ð1x2 ÞP00n ðxÞ2xPn0 ðxÞþnðnþ1ÞPn ðxÞ ¼ 0: ð151Þ
Multiplying (150) by Pn(x) and (151) by Pm(x) and
subtracting, we get
2n þ 1
2
Pm ðxÞPn ðxÞdx ¼ 0; m 6¼ n:
1
If remains to show that
R1
1
2
P2n ðxÞdx ¼ 2nþ1
.
In this direction, we have by generating functions,
ð1 x2 Þ½Pn ðxÞP00m ðxÞ Pm ðxÞP00n ðxÞ
ð1 2xt þ t2 Þ 2 ¼
1
2x½Pn ðxÞP0m ðxÞ þ Pm ðxÞP0n ðxÞ
¼ ðn2 m2 þ n mÞPm ðxÞPn ðxÞ
Z1
and
1
d
½ð1 x2 ÞðPn ðxÞP0m ðxÞ Pm ðxÞP0n ðxÞÞ
dx
¼ ðn mÞðn þ m þ 1ÞPm ðxÞPn ðxÞ:
Integration on the interval [–1, 1] yields
½ð1 x2 ÞðPn ðxÞP0m ðxÞ Pm ðxÞP0n ðxÞ11
Z1
¼ ðn mÞðn þ m þ 1Þ
Pm ðxÞPn ðxÞdx
t m Pm ðxÞ
m¼0
ð1 2xt þ t2 Þ 2 ¼
or
1
X
1
X
t n Pn ðxÞ:
n¼0
Multiplying these two expressions, we get
ð1 2xt þ t2 Þ1 ¼
1 X
1
X
t mþn Pm ðxÞPn ðxÞ:
m¼0 n¼0
Integrating w.r.t. x on [– 1, 1], we have
Z1
1
ð1 2xt þ t2 Þ1 dx
1
or
Z1
0 ¼ ðn mÞðn þ m þ 1Þ
Pm ðxÞPn ðxÞdx
1
¼
1
1 X
1 Z
X
m¼0 n¼0
1
t mþn Pm ðxÞPn ðxÞdx:
Ordinary Differential Equations
Since
R1
1
Pm ðxÞPn ðxÞdx ¼ 0 for m 6¼ n, the above
n¼0
1
½1680x4 1440x2 þ 144
384
1
¼ ½35x4 30x2 þ 3:
8
1
1 Z
X
n¼0
t2n P2n ðxÞdx ¼ 1
P4 ðxÞ ¼
1
Thus
1
logð1 þ t2 2xt
2t
1
x¼1
2
1
ð1 tÞ
1
1þt
¼ log
¼ log
2t
1t
ð1 þ tÞ2 t
3
5
1
t
t
¼ 2 t þ þ þ ...
3 5
t
t2 t4
t2n
¼ 2 1 þ þ þ ... þ
þ ... :
3 5
2n þ 1
Comparing the coefficients of t2n, we have
Z1
2
P2n ðxÞdx ¼
2n þ 1
EXAMPLE 1.170
Establish the following recurrence relations for
Legendre polynomial.
(i) ðn þ 1ÞPnþ1 ðxÞ ¼ ð2n þ 1ÞxPn ðxÞ nPn1 ðxÞ
0
0
(ii) nPn ðxÞ ¼ xPn ðxÞ Pn1 ðxÞ
0
0
(iii) ð2n þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞ Pn1 ðxÞ
0
0
(iv) ðn þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞ xPn ðxÞ
0
(v) ð1 x2 ÞPn ðxÞ ¼ nðPn1 ðxÞ xPn ðxÞÞ
0
(vi) ð1 x2 ÞPn ðxÞ ¼ ðn þ 1ÞðxPn ðxÞ Pnþ1 Þ.
Solution. (i) The generating formula for Pn(x) is
1
X
1
tn Pn ðxÞ:
ð1 2xt þ t2 Þ 2 ¼
n¼0
1
or
2n þ 1
2
1.93
Therefore,
expression reduces to
1
Z1
1 Z
X
2 1
ð1 2xt þ t Þ dx ¼
t2n P2n ðxÞdx:
1
n
Differentiating w.r.t. t, we get
Z1
P2n ðxÞdx ¼ 1:
1
EXAMPLE 1.169
Using Rodrigue’s formula, find the polynomial
expression for P4 ðxÞ.
Solution. The Rodrigue’s formula is
1 dn 2
ðx 1Þn :
Pn ðxÞ ¼ n
2 n! dxn
Therefore,
1 d4 2
1 d4 2
ðx 1Þ4 ¼
ðx 1Þ4 :
P4 ðxÞ ¼ n
4
2 4! dx
384 dx4
So let
y ¼ ðx2 1Þ4 ¼ x8 4x6 þ 6x4 4x2 þ 1:
Then,
dy
¼ 8x7 24x5 þ 24x3 8x
dx
d2y
¼ 56x6 120x4 þ 72x2 8
dx2
d3y
¼ 336x5 480x3 þ 144x
dx3
d4y
¼ 1680x4 1440x2 þ 144:
dx4
1
X
3
1
nPn ðxÞtn1
ð1 2xt þ t2 Þ 2 ð2x þ 2tÞ ¼
2
n¼0
or
ðxtÞð12xt þt2 Þ 2 ¼ ð12xt þt2 Þ
1
1
X
nPn ðxÞtn1
n¼0
or
ðxtÞ
1
X
tn Pn ðxÞ ¼ ð12xt þt2 Þ
n¼0
1
X
ntn1 Pn ðxÞ:
n¼0
n
Comparing coefficients of t , we get
xPn ðxÞ Pn1 ðxÞ ¼ ðn þ 1ÞPnþ1 ðxÞ 2nxPn ðxÞ
þ ðn 1ÞPn1 ðxÞ
or
ðn þ 1ÞPnþ1 ðxÞ ¼ ð2n þ 1ÞxPn ðxÞ nPn1 ðxÞ
(ii) Differentiating the generating formula w.r.t. x,
we get
1
X
3
1
0
tn Pn ðxÞ
ð1 2xt þ t2 Þ 2 ð2tÞ ¼
2
n¼0
1.94
n
Engineering Mathematics-II
or
(v) From part (iv), we have
1
X
tð1 2xt þ t2 Þ 2 ¼
3
0
0
tn Pn ðxÞ:
n¼0
Replacing n by n – 1, we get
On the other hand, differentiation of the generating
formula w.r.t. t yields
1
X
3
ntn1 Pn ðxÞ: ð153Þ
ðx tÞð1 2tx þ t2 Þ 2 ¼
0
0
nPn1 ðxÞ ¼ Pn ðxÞ xPn1 ðxÞ:
ð155Þ
Also, from part (ii), we have
0
0
nPn ðxÞ ¼ xPn ðxÞ Pn1 ðxÞ
n¼0
or
Dividing (153) by (152), we get
0
0
nxPn ðxÞ ¼ x2 Pn ðxÞ xPn1 ðxÞ: ð156Þ
1
P
ntn1 Pn ðxÞ
x t n¼0
¼ P
1
t
tn P0n ðxÞ
Subtracting (156) from (155), we get
0
nPn1 ðxÞ nxPn ðxÞ ¼ ð1 x2 ÞPn ðxÞ
n¼0
or
or
1
X
0
ðn þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞ xPn ðxÞ:
ð152Þ
nt Pn ðxÞ ¼ ðx tÞ
n
n¼0
1
X
0
ð1 x2 ÞPn ðxÞ ¼ n½Pn1 ðxÞ xPn ðxÞ:
0
t Pn ðxÞ:
n
(vi) From (155), we have
n¼0
n
Comparing the coefficients of t on both sides, we
have
0
0
nPn ðxÞ ¼ xPn ðxÞ Pn1 ðxÞ:
0
0
nxPn1 ðxÞ ¼ xPn ðxÞ x2 Pn1 ðxÞ
ð157Þ
and from part (ii), we have
0
0
nPn ðxÞ ¼ xPn ðxÞ Pn1 ðxÞ: ð158Þ
(iii) From (i), we have
ðn þ 1ÞPnþ1 ðxÞ ¼ ð2n þ 1ÞxPn ðxÞ nPn1 ðxÞ:
Subtracting (158) from (157), we obtain
Differentiating w.r.t. x, we get
ðn þ 1ÞP0nþ1 ðxÞ ¼ ð2n þ 1ÞPn ðxÞ
Replacing n by n + 1, we get
0
þ ð2n þ 1ÞxP0n ðxÞ nP0n1 ðxÞ:
ð154Þ
ð1 x2 ÞPn ðxÞ ¼ ðn þ 1Þ½xPn ðxÞ Pnþ1 ðxÞ :
EXAMPLE 1.171
Show that if n is odd, then
But from part (ii)
0
0
n½xPn1 ðxÞ Pn ðxÞ ¼ ð1 x2 ÞPn1 ðxÞ :
0
xPn ðxÞ ¼ nPn ðxÞ þ Pn1 ðxÞ:
0
Pn ðxÞ ¼ ð2n 1ÞPn1 ðxÞ þ ð2n 5ÞPn3 ðxÞ
Therefore, (154) reduces to
þ ð2n 9ÞPn5 ðxÞ þ . . . þ 3P1 ðxÞ :
0
ðn þ 1ÞPnþ1 ðxÞ ¼ ð2n þ 1ÞPn ðxÞ
0
0
þ ð2n þ 1Þx½nPn ðxÞ þ Pn1 ðxÞ nPn1 ðxÞ
Solution. We have
0
0
ð2n þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞPn1 ðxÞ
or
ð2n þ 1ÞPn ðxÞ ¼ P0nþ1 P0n1 :
Changing n to (n – 1), (n – 3), (n – 5),. . ., we get
(iv) From part (ii), we have
0
0
ð2n þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞ Pn1 ðxÞ
and from part (ii), we have
0
0
nPn ðxÞ ¼ xPn ðxÞ Pn1 ðxÞ:
Subtracting, we obtain
0
ðrecurrence formulaÞ:
0
ðn þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞ xPn ðxÞ:
0
0
ð2n 1ÞPn1 ðxÞ ¼ Pn ðxÞ Pn2 ðxÞ
0
0
ð2n 5ÞPn3 ðxÞ ¼ Pn2 ðxÞ Pn4 ðxÞ
0
0
ð2n 9ÞPn5 ðxÞ ¼ Pn4 ðxÞ Pn6 ðxÞ
...
...
0
0
3P1 ðxÞ ¼ P2 ðxÞ P0 ðxÞ if n is odd:
Ordinary Differential Equations
Adding these equations, we get
ð2n 1ÞPn1 ðxÞ þ ð2n 5ÞPn3 ðxÞ
þ ð2n 9ÞPn5 ðxÞ þ . . . þ 3P1 ðxÞ
0
0
0
¼ Pn ðxÞ P0 ðxÞ ¼ Pn ðxÞ 0
0
¼ Pn ðxÞ; since P0 ðxÞ ¼ 1:
n
1.95
Solution. (i) Generating function formula is
1
X
1
tn Pn ðxÞ :
ð1 2xt þ t2 Þ 2 ¼
n¼0
Therefore,
1
X
1
tn Pn ðxÞ ¼ ð1þ2xt þt2 Þ 2
n¼0
¼ ½12xðtÞþt2 2
1
1
X
X
¼
ðtÞn Pn ðxÞ ¼
ð1Þn tn Pn ðxÞ
1
EXAMPLE 1.172
Show that
Z1
xPn ðxÞPn1 ðxÞdx ¼
1
2n
:
4n2 1
Solution. We know that
ðn þ 1ÞPnþ1 ðxÞ ¼ ð2n þ 1ÞxPn ðxÞ nPn1 ðxÞ :
n¼0
n¼0
and so Pn(–x) ¼ (–1)n Pn (x) .
(ii) We have
1
X
1
Pn ð1Þtn ¼ ð1 2t þ t2 Þ 2 ¼ ð1 tÞ1
n¼0
¼ 1 þ t þ t2 þ . . . þ t n þ . . .
1
X
tn :
¼
Changing n to n –1, we get
nPn ðxÞ ¼ ð2n 1ÞxPn1 ðxÞ ðn 1ÞPn2 ðxÞ
n¼0
or
xPn1 ðxÞ ¼
1
½nPn ðxÞ þ ðn 1ÞPn2 ðxÞ :
2n 1
Therefore,
Hence Pn ð1Þ ¼ 1 :
(iii) We note that
1
X
1
tn Pn ð1Þ ¼ ð1 þ 2t þ t2 Þ 2
Z1
n¼0
¼ ð1 þ tÞ1 ¼
xPn ðxÞPn1 ðxÞdx
1
2n1
n
¼
2n1
¼
ð1Þn tn :
n¼0
1
¼
1
X
1
2n1
Z1
½nPn ðxÞþðn1ÞPn2 ðxÞPn ðxÞdx
1
Z1
P2n ðxÞdxþ
1
n1
2n1
Z1
Pn ðxÞPn2 ðxÞdx
1
Z1
P2n ðxÞdxþ0 using orthogonal property
1
n
2
2n
¼
:
¼ 2
2n1 2nþ1
4n 1
EXAMPLE 1.173
Show that
(i) Pn ðxÞ ¼ ð1Þn Pn ðxÞ
(ii) Pn ð1Þ ¼ 1
(iii) Pn ð1Þ ¼ ð1Þn :
Hence Pn ð1Þ ¼ ð1Þn :
Second Method. From part (i)
Pn ðxÞ ¼ ð1Þn Pn ðxÞ :
Therefore,
Pn ð1Þ ¼ ð1Þn Pn ð1Þ ¼ ð1Þn usingðiiÞ:
1.27
FOURIER–LEGENDRE EXPANSION OF A
FUNCTION
We have
qffiffiffiffiffiffiffiffiestablished in Examples 1.167 and 1.168
that 2nþ1
2 fPn ðxÞg constitutes an orthonormal system. Therefore, as in the case of Fourier expansion,
the expansion in terms of Pn(x) of a given function
f (x) is possible. Let
1
X
cn Pn ðxÞ :
ð159Þ
fðxÞ ¼
n¼0
1.96
n
Engineering Mathematics-II
Multiplying both sides by Pn(x) and integrating both
sides by [–1, 1], we get
Z1
Z1
2
2
f ðxÞPn ðxÞdx ¼ cn
Pn ðxÞdx ¼ cn
2n þ 1
1
and so on. Hence, Fourier–Legendre expansion of
the given function is
f ðxÞ ¼
1
3
7
27
¼ P0 ðxÞ þ P1 ðxÞ P3 ðxÞ þ P4 ðxÞ
2
4
16
16
þ ...
and so
Z1
f ðxÞPn ðxÞdx :
1
The coefficients cn are called Fourier–Legendre
coefficients and the expansion (159) is called
Fourier–Legendre expansion of f (x) .
1.28
x dy y dx ¼ ðx2 þ y2 Þ dx
0;
1;
f ðxÞ ¼
1 < x < 0
0 < x < 1:
Solution. We have
x dy y dx ¼ ðx2 þ y2 Þdx
in Fourier–Legendre series.
or
Solution. We have
2n þ 1
2
Z1
f ðxÞPn ðxÞdx ¼
1
2n þ 1
2
x dy y dx
¼ dx
x2 þ y2
Z1
Pn ðxÞdx: or
1
Therefore,
1
c0 ¼
2
c1 ¼
c2 ¼
3
2
5
2
Z1
1
P0 ðxÞdx ¼
2
Z1
0
0
Z1
Z1
P1 ðxÞdx ¼
3
2
0
0
Z1
P2 ðxÞdx ¼
5
2
Z1
1
dx ¼ ;
2
xdx ¼
Z1
0
7
c3 ¼
2
3 1
3
¼ ;
2 2
4
7
P3 ðxÞdx ¼
2
Z1
y
tan1 ¼ x þ C:
x
EXAMPLE 1.176
1
Solve ð1 þ y2 Þdx ¼ ðetan y xÞdy.
5x3 3x
dx
2
which is of the form
7 x4
x2
7
¼ 5 3
¼ ;
2 0
4 4
16
Z1
Z1
9
9 35x4 30x2 þ 3
27
c4 ¼
dx ¼ ;
P4 ðxÞdx ¼
2
2
8
16
0
Integrating, we get
Solution. The given equation can be written as
0
1
y
d tan1
¼ dx:
x
3x2 1
5
dx ¼ ½x3 x10 ¼ 0;
2
4
0
0
0
MISCELLANEOUS EXAMPLES
EXAMPLE 1.175
Solve the differential equation
EXAMPLE 1.174
Expand
cn ¼
cn Pn ðxÞ
n¼0
1
2n þ 1
cn ¼
2
1
X
1
dx
x
etan y
¼
;
þ
2
1 þ y2
dy 1 þ y
dx
þ Px ¼ Q;
dy
where P and Q are function of y only. The integrating factor is
R 1
1
dy
I:F ¼ e 1þy2 ¼ etan y :
Ordinary Differential Equations
Therefore the solution of the differential equation is
Z tan1 y
e
1
tan1 y
¼
:etan y dy þ c
xe
1 þ y2
Z 2tan1 y
e
e2 tan 1 y
dy
þ
c
¼
¼
þc
1 þ y2
2
or
2xetan
1
y
1
¼ e2tan
y
þ c1 :
EXAMPLE 1.177
Solve the initial value problem ex ðcos ydx
sin ydyÞ ¼ 0, yð0Þ ¼ 0.
y constant
or
Therefore
@M
¼ ex sin y
@y
or
@N
¼ ex sin y:
@x
Hence the given equation is exact. The solution of
the given differential equation is
Z
Z
Mdx þ ðterms in N not containing xÞdy ¼ k
yconstant
or
ex cos y ¼ k:
Using the initial condition yð0Þ ¼ 0, we get
k ¼ e0 cos 0 ¼ 1. Hence the required solution is
ex cos y ¼ 1:
EXAMPLE 1.178
Solve the1 following equation
ðxy2 ex3 Þdx x2 y dy ¼ 0:
Solution. The equation here is
1
xy2 ex3 dx x2 ydy ¼ 0
Comparing it with Mdx þ Ndy ¼ 0, we have
M ¼ xy2 e
Therefore
@M
¼ 2xy
@y
1
x3
and
and
N ¼ x2 y
@N
¼ 2xy:
@x
1.97
@N
Since @M
@y 6¼ @x ; the given equation is not exact. But
@M
@N
2xy ð2xyÞ
4
@y @x
¼
¼ ;
2
N
x y
x R
4
which is a function of x only. Hence e x dx ¼
e4 log x ¼ x14 is the integrating factor. So multiplying
the given equation throughout by x14 , we get
y2 1 1
y
ð 3 4 ex3 Þdx 2 dy ¼ 0;
x
x
x
which is an exact equation. Therefore the required
solution is
2
Z
y
1 13
x
dx ¼ c
4e
3
x
y constant x
Solution. The given equation is of the form
Mdx þ Ndy ¼ 0, where
M ¼ ex cos y and N ¼ ex sin y:
and
n
y2
þ
2x2
Z
1 13 3
dx ¼ c
ex
3
x4
y2
1 1
þ ex3 ¼ c:
2
2x
3
EXAMPLE 1.179
Solve the differential equation
ðy cos x þ 1Þ dx þ sin x dy ¼ 0:
Solution. The given differential equation is of the
form Mdx þ Ndy ¼ 0, where
M ¼ y cos þ1; N ¼ sin x:
Now
@M
@N
¼ cos x;
¼ cos x:
@y
@x
Therefore the given equation is exact and its solution is
Z
Z
ðy cos þ1Þdx þ 0dy ¼ C
y constant
or
y sin x þ x ¼ C:
EXAMPLE 1.180
If the air is maintained at 30 C and the temperature
of the body cools from 80 C to 60 C in 12 minutes,
find the temperature of the body after 24 minutes.
1.98
n
Engineering Mathematics-II
Solution. Let T be the temperature of the body at any
instant t. By Newton’s law of cooling, we have
dT
¼ kðT T0 Þ
dt
and so variable separation yields
dT
¼ kdt:
T T0
Integrating, we get
logðT T0 Þ ¼ kt þ log C
or
dT
¼ k dt:
dt
Integrating, we get
logðT T0 Þ ¼ kt þ log C
or
T T0 ¼ Cekt :
or
T 30 ¼ c ekt :
Therefore
C¼
But when t ¼ 10 ; T ¼ 250 . Therefore
250 ¼ 80 þ 220 e10k ;
which yields
T 30 ¼ 50 ekt :
e10k ¼
Now when t ¼ 12 ; T ¼ 60 . Therefore
60 30 ¼ 50 e12k
k¼
30 ¼ 50 e12 k
1
5
k ¼ log :
12
3
T ¼ 30 þ 50 eð12 log3Þ t
1
1
29
log ¼ 0:02578:
10
7
Hence
or
which yields
250 80 17
¼
220
22
or
or
3
12k ¼ log
5
Hence
ð160Þ
Equation (160) gives
T 80 ¼ C e kt
When t ¼ 0 ; T ¼ 300 . Therefore C ¼ 300
80 ¼ 220 . Thus
T ¼ 80 þ 220 ekt :
T T0 ¼ C ekt
But when t ¼ 0 ; T ¼ 80 :
80 30 ¼ 50 and we have
or
T ¼ 80 þ 220 e0:2578t :
Therefore temperature of the body after 20 minutes is
T ¼ 80 þ 220 e5:156 :
5
When t ¼ 24, we have
T ¼ 30 þ 50 e2 log3
5
9
¼ 30 þ 50 elog25
9
¼ 30 þ 50
¼ 30 þ 18 ¼ 48 C:
25
EXAMPLE 1.181
Suppose that an object is heated to 300 F and allowed
to cool in a room whose air temperature is 80 F. If
after 10 minutes the temperature of the object is
250 F, what will be its temperature after 20 minutes?
Solution. By Newton’s law of cooling, we have
dT
¼ kðT T0 Þ;
dt
EXAMPLE 1.182
A bacterial culture growing exponentially increases
200 to 500 grams in the period from 6 AM to 9 AM.
How many grams will it be at noon?
Solution. The governing differential equation of this
problem is
dx
¼ kx;
dt
which has the solution as
ð161Þ
x ¼ C ekt :
When t ¼ 0 ; x ¼ 200. Therefore (161) gives
C ¼ 200. Thus
ð162Þ
x ¼ 200 ekt :
When t ¼ 3 hours, x ¼ 500. Therefore (162) yields
500 ¼ 200 e3k
Ordinary Differential Equations
or
n
1.99
or
5
ð163Þ
e3k ¼
2
At noon t ¼ 6 hour, using (163), we have
2
5
6k
3k 2
x ¼ 200 e ¼ 200ðe Þ ¼ 200
¼ 1250 grams:
2
log r þ 2 log cos
h
¼ log c
2
or
h
log r þ log cos2 ¼ log c
2
or
h
rcos2 ¼ c
2
EXAMPLE 1.183
Find the orthogonal trajectories of the family
2a
¼ 1 cos h:
r
or
r
ð1 þ cos hÞ ¼ c
2
or
Solution. We have
2c
;
1 þ cos h
which is the required orthogonal trajectory.
r¼
2a
¼ 1 cos h;
r
or
2a
:
ð164Þ
1 cos h
Differentiating (164) with respect to h; we get
dr ð1 cos hÞ ð0Þ 2aðsin hÞ
¼
dh
ð1 cos hÞ2
r¼
¼
2a sin h
ð1 cos hÞ2
:
ð165Þ
r
dh
h
¼ cot
dr
2
or
sin h2
dr
h
¼ tan dh ¼
dh:
r
2
cos h2
Integrating, we get
h
log r ¼ 2 log cos þ log c
2
or
which yields m ¼ 1; 2; 3. Therefore
C.F: ¼ c1 ex þ c2 e2x þ c3 e3x :
2 sin h2 cos h2
1 dr
sin h
¼
¼
r dh
1 cos h
2sin2 h2
or
Solution. The auxiliary equation for the given nonhomogeneous equation is
m3 6m2 þ 11m 6 ¼ 0
ðm 1Þðm 2Þðm 3Þ ¼ 0;
Dividing (165) by (164), we get
h
¼ cot :
2
dr
in (166), we get
Replacing dh by r2 dh
dr 1
dh
h
r2
¼ cot
r
dr
2
EXAMPLE.1.184
Solve ðD3 6D2 þ 11D 6Þy ¼ e2x þ e3x
ð166Þ
The particular integral is given by
1
e2x
P.I ¼
ðD 1ÞðD 2ÞðD 3Þ
1
e3x
þ
ðD 1ÞðD 2ÞðD 3Þ
1
e2x
¼
ð2 1Þð2 2Þð2 3Þ
1
þ
e3x :
ð3 1Þð3 2Þð3 3Þ
1 2x
1 3x
¼ e e :
60
120
Hence the complete solution of the given differential equation is
y ¼ C.F: þ P.I
¼ C1 ex þ C2 e2x þ c3 e3x 1 2x
1 3x
e e :
60
120
1.100
n
Engineering Mathematics-II
EXAMPLE.1.185
Solve ðD2 þ 4D þ 13Þy ¼ e2x
Hence the complete solution is
Solution. The auxiliary equation for the given
differential equation is
m2 4m þ 13 ¼ 0;
pffiffiffiffiffiffiffiffiffiffi
which yields m ¼ 4 1652
¼ 2 3i. Therefore
2
C.F ¼ ðc1 cos 3x þ c2 sin 3xÞe2x :
1
y ¼ C.F þ P.I ¼ c1 cos x þ c2 sin x þ x sin x
4
1
1 x 2
þ cos 3x þ e ðx 2x þ 1Þ:
16
2
EXAMPLE 1.187
Solve ðD2 þ 1Þy ¼ sin2 x.
Solution. We have
The particular integral is given by
1
1
e2x
FðxÞ ¼ 2
P.I ¼
f ðDÞ
D 4D þ 13
1
1
e2x ¼ e2x :
¼ 2
2 4ð2Þ þ 13
9
The auxiliary equation is m2 þ 1 ¼ 0; which yields
m ¼ i. Therefore
C.F: ¼ c1 cos x þ c2 sin x:
Hence the complete solution is
Now
1
y ¼ C.F þ P.I ¼ ðc1 cos 3x þ c2 sin 3xÞe2x þ e2x :
9
EXAMPLE.1.186
Solve ðD2 þ 1Þy ¼ sin x sin 2x þ ex x2 .
Solution. The auxiliary equation for the given differential equation is
m2 þ 1 ¼ 0;
which yields
m¼
2:
Therefore
C.F ¼ c1 cos x þ c2 sin x:
The particular integral is given by
1
1
sin x sin 2x þ 2
ex x2
P.I ¼ 2
D þ1
D þ1
1
1
1
ex x2
¼ 2
ðcos x cos 3xÞ þ 2
D þ1 2
D þ1
1
1
cos x cos 3x
¼
2ðD2 þ 1Þ
2ðD2 þ 1Þ
1
þ ex
x2
ðD þ !Þ2 þ 1
x
1
cos 3x
¼ sin x 4
2ð9 þ 1Þ
1
ex
D2
þ
1þDþ
x2
2
2
x
1
1
¼ sin x þ cos 3x þ ex ðx2 2x þ 1Þ:
4
16
2
ðD2 þ 1Þy ¼ sin2 x:
1
1
1 cos 2x
FðxÞ ¼ 2
f ðDÞ
D þ1
2
1
½1 cos 2x
¼
2ðD2 þ 1Þ
1
1
1
1
¼ cos 2x ¼ cos 2x
2 2ðD2 þ 1Þ
2 2ð4 þ 1Þ
1 1
¼ þ cos 2x:
2 6
Hence the solution is
1 1
y ¼ c1 cos x þ c2 sin x þ þ cos 2x:
2 6
P.I ¼
EXAMPLE 1.188
Find the particular integral of ðD2 2D þ lÞy ¼
cosh x:
Solution. The required particular integral is
1
cosh x
P.I ¼ 2
D 2D þ 1
1
ex þ ex
¼ 2
2
D 2D þ 1
1
1
1
1
ex þ : 2
ex
¼ : 2
2 D 2D þ 1
2 D 2D þ 1
1
1
1
ex þ ex
¼ x:
2 2D 2
8
1 2 1 x 1 x
¼ x : e þ e
2
2
8
1 2 x 1 x
¼ x e þ e :
4
8
Ordinary Differential Equations
EXAMPLE 1.189
Solve the differential equation
Solution. The auxiliary equation for the given differential equation is
m2 þ 4 ¼ 0;
2i: Therefore,
C.F ¼ c1 cos 2x þ c2 sin 2x:
Further,
1
1
P.I ¼ 2
ðxsin xÞ ¼ 2
x½I.P: of eix D þ4
D þ4
"
#
1
1
ix
ix
x e ¼ I.P: of e
¼ I.P of 2
x
D þ4
ðD þ iÞ2 þ 4
¼ I.P: of eix
1
D2 þ 2iD 1 þ 4
1.101
Hence the complete solution of the given differential equation is
d2y
þ 4y ¼ x sin x:
dx2
which yields m ¼
n
x
1
x
¼ I.P: of eix 2
D þ 2iD 1 þ 3
"
#
1
x
¼ I.P: of eix 2
2
3 1 þ 3 iD þ D3
" 1 #
1
2
D2
ix
¼ I.P: of e 1 þ iD þ
x
3
3
3
1
2
D2
¼ I.P of eix 1 iD 3
3
3
# )
2 2
2
D
þ iD þ
x
3
3
1
2
¼ I.P of eix x i
3
3
1
2
¼ I.P: of ðcos x þ isin xÞ x i
3
3
1
2
2
¼ I.P: of xcos x þ ix sin x i cos þ sin x
3
3
3
1
2
2
¼ I.P: of xcos x þ sin x þ i x sin x cos x
3
3
3
1
2
¼ x sin x cos x :
3
3
y ¼ c1 cos 2x þ c2 sin 2x þ
1
2
x sin x cos x :
3
3
EXAMPLE 1.190
2
3x
Solve: ddx2y 3 dy
dx þ 2y ¼ xe þ sin 2x:
Solution. The auxiliary equation for the given differential equation is m2 3m þ 2 ¼ 0, which
yields m ¼ 1; 2. Hence
C.F: ¼ c1 ex þ c2 e2x :
Further,
P.I ¼
¼
1
1
½xe3x þ sin 2x
FðxÞ ¼ 2
f ðDÞ
D 3D þ 2
1
1
ðxe3x Þ þ 2
ðsin 2xÞ
D2 3D þ 2
D 3D þ 2
¼ e3x
þ
1
1
2
ðD þ 3Þ 3ðD þ 3Þ þ 2
x
1
ðsin 2xÞ
4 3D þ 2
2 3D
ðsin 2xÞ
ð2 þ 3DÞð2 3DÞ
1
1 3x
D2 þ 3D
2 3D
¼ e 1þ
x
ðsin 2xÞ
2
2
4 9D2
1 3x
3D D2
2 3D
¼ e 1
þ x ðsin 2xÞ
2
2
2
4 þ 36
1 3x
3
1
½2 sin 2x 3Dðsin 2xÞ
¼ e x
2
2
40
¼ e3x
1
D2 þ 3D þ 2
x
1
3
1
6
¼ xe3x e3x sin 2x þ cos 2x
2
4
20
40
1
1
¼ e3x ð2x 3Þ þ ð3 cos 2x sin 2xÞ:
4
20
Hence the complete solution of the given differential equation is
1
1
y ¼ c1 ex þ c2 e2x þ e3x ð2x 3Þ þ ð3 cos 2x
4
20
sin 2xÞ:
1.102
n
Engineering Mathematics-II
EXAMPLE 1.191
Solve:
d2 x
dx
dt2 4 dt þ 13x ¼ 0 with xð0Þ ¼ 0;
dxð0Þ
dt
¼ 2:
Solution. The auxiliary equation for the given equation is
m2 4m þ 13 ¼ 0;
which yield m ¼ 2
solution is
3i. Therefore the general
x ¼ e2t ½c1 cos 3t þ c2 sin 3t:
When t ¼ 0 we have x ¼ 0. Therefore c1 ¼ 0.
Further,
dx
¼ e2t ½3c1 sin 3t þ 3c2 cos 3t
dt
þ 2e2t ½c1 cos 3t þ c2 sin 3t
When t ¼ 0;
dx
dt
¼ 2 and so we have
2
2 ¼ 3c2 þ 2c1 ) c2 ¼ :
3
Hence
2
x ¼ e2t sin 3t:
3
EXAMPLE 1.192
Solve ðx2 D2 þ 4xD þ 2Þy ¼ x log x.
Solution. The given Cauchy–Euler equation is
x2
d2y
dy
þ 4x þ 2y ¼ x log x:
dx2
dx
Proceeding as in Example 1.115, we put x ¼ et and
reduce the equation to the form
½DðD 1Þ þ 4D þ 2y ¼ t et :
The C.F for this equation is
C.F ¼ c1 et þ c2 e2t :
The particular integral is
1
t et
P.I ¼ 2
D þ 3D þ 2
1
¼ et
t
2
ðD þ 1Þ þ 3ðD þ 1Þ þ 2
1
t
¼ et 2
D þ 5D þ 6
1
t
¼ et 6 1 þ 56 D þ 16 D2
1
1
5
1
1 þ D þ D2
¼ et
t
6
6
6
1
5
1
¼ et 1 D D 2 þ t
6
6
6
1 t
5
1 t
5
¼ t e et :
¼ e t
6
6
6
36
Hence the solution is
1
5
y ¼ c1 et þ c2 e2t þ t et et
6
36
c1 c2 1
5
¼ þ 2 þ x log x x:
x x
6
6
EXAMPLE 1.193
Convert ðx3 D2 þ 3x2 D þ 5xÞy ¼ 2 into a differential equation with constant co-efficients.
Solution. Given equation can be written as
2
ðx2 D2 þ 3xD þ 5Þy ¼ ;
x
which is Cauchy’s homogeneous linear difference
equation,
Let x ¼ et : Then, we have x dy
dx ¼ Dy and
2
d
y
x2 dx2 ¼ DðD 1Þy; where D ¼ dtd . Hence given
equation reduces to
½DðD 1Þ þ 3D þ 5y ¼ 2:et or
ðD2 þ 2D þ 5Þy ¼ 2et :
EXAMPLE 1.194
Solve
ðxD2 þ DÞy 0:
Solution. The given differential equation with variable coefficients is
x D2 y þ Dy ¼ 0:
Ordinary Differential Equations
x D y þ x Dy ¼ 0;
2
which is Cauchy–Euler equation. Putting x ¼ et so
that t ¼ log x, we have
dy
d2y
¼ Dy and x2 2 ¼ DðD 1Þy:
dx
dx
Consequently, the given equation transforms to
½DðD 1Þ þ D y ¼ 0 or D2 y ¼ 0. Integrating
twice, we get
y ¼ at þ b ¼ a log x þ b:
x
ðx2 D2 xD 3Þ y ¼ x2 ðlog xÞ2
Solution. The given equation
x2
d2y
dy
x 3y ¼ x2 ðlog xÞ2
2
dx
dx
is a Cauchy–Euler homogeneous linear equation.
So put x ¼ et so that t ¼ log x. Then
x
dy
¼ Dy
dx
and
x2
d2y
¼ DðD 1Þy;
dx2
EXAMPLE 1.195
Solve ðx2 D2 3xD þ 4Þy ¼ x2 given that yð1Þ ¼ 1
and y0 ð1Þ ¼ 0.
and the given equation transforms to
Solution. We are given the differential equation
or
ðx2 D2 3x D þ 4Þ y ¼ x2 ;
which is a Cauchy–Euler homogeneous equation.
Putting x ¼ et so that t ¼ log x, we have
dy
d2y
¼ Dy and x2 2 ¼ ½DðD 1Þy:
dx
dx
Therefore the given equation transforms to
½DðD 1Þ D 3y ¼ t2 e2t
ðD2 2D 3Þy ¼ t2 e2t :
The roots of the auxiliary equation m2 2m 3 ¼ 0
are m ¼ 3 and m ¼ 1. Therefore
x
C.F ¼ c1 e3t þ c2 et :
Now
½DðD 1Þ 3D þ 4y ¼ e2t
or
D2 4D þ 4 y ¼ e2t
which yields
ð167Þ
m ¼ 2; 2:
Therefore
C.F: ¼ ðc1 þ c2 tÞe2t :
Further
1
e2t
D2 4D þ 4
1
1
¼t
e2t ¼ t2 : e2t
2D 4
2
1 2 2t
¼ t e :
2
Hence the required solution is
P: I ¼
y ¼ C: F þ P: I
¼ ðc1 þ c2 log xÞ x2 þ
x2
ðlog xÞ2 :
2
1
ðt2 e2t Þ
D2 2D 3
1
¼ e2t
t2
2
½ðD þ 2Þ 2ðD þ 2Þ 3
1
2t
t2
¼e 2
D þ 2D 3
e2t
1
t2
¼ 2
3 1 3 D D32
1
1
2D D2
¼ e2t 1 t2
þ
3
3
3
1
2D D2 4D2
þ
þ t2
¼ e2t 1 þ
þ
3
9
3
3
1
2D 7D2
þ t2
¼ e2t 1 þ
þ
9
3
3
1
2
7
¼ e2t t2 þ Dðt2 Þ þ D2 ðt2 Þ
3
3
9
1
4
14
:
¼ e2t t2 þ t þ
3
3
9
P.I ¼
The auxiliary equation for the equation (167) is
m2 4m þ 4 ¼ 0;
1.103
EXAMPLE 1.196
Solve the differential equations:
Multiplying throughout by x, we get
2
n
1.104
n
Engineering Mathematics-II
Therefore the complete solution is
y ¼ C.F þ P.I
Hence the complete solution of the given differential equation is
1
1
3
y ¼ C.F þ P.I ¼ c1 e4t þ c2 et e2t t þ
6
2
8
2
c
x
1
3
2
¼ c1 x4 þ log x þ :
x
6 2
8
1
4
14
¼ c1 e3t þ c2 et e2t t2 þ t þ
3
3
9
2
c2 x
4
14
ðlog xÞ2 þ log x þ
:
¼ c1 x3 þ x
3
3
9
EXAMPLE 1.197
Solve the differential equation
x2 d 2 y
dy
2x 4y ¼ x2 þ 2 log x:
dx2
dx
EXAMPLE 1.198
2
Solve: ð2x þ 3Þ2 ddx2y þ 5ð2x þ 3Þ dy
dx þ y ¼ 4x:
Solution. We have
ð2x þ 3Þ2
Solution. The given Cauchy–Euler homogeneous
linear equation is
d2y
dy
x2 2 2x 4y ¼ x2 þ 2 log x:
dx
dx
Substitute x ¼ et so that t ¼ log x. Then
dy
d2y
¼ DðD 1Þy:
x ¼ Dy and
dx
dx2
Therefore the given equation reduces to
DðD 1Þy 2Dy 4y ¼ e þ 2t
d2y
dy
þ 5ð2x þ 3Þ þ y ¼ 4x;
dx2
dx
which is Legendre’s linear equation. So putting
2x þ 3 ¼ et or t ¼ logð2x þ 3Þ, the given equation
reduces to
et 3
4ðD2 DÞ þ 10D þ 1 y ¼ 4
2
or
ð4D2 þ 6D þ 1Þy ¼ 2et 6:
2t
or
ðD2 3D 4Þy ¼ e2t þ 2t:
The characteristic equation for the equation is
m2 3m 4 ¼ 0;
which yields
m ¼ 4; 1:
Thus
C.F ¼ c1 e4t þ c2 et :
The particular integral is given by
1
2t
e2t þ 2
et :
P.I ¼ 2
D 3D þ 4
D 3D þ 4
e2t
2t
¼
4 6 4 4 1 D3D
4
1
e2t 1
D2 3D
1
t
¼ 6 2
4
1 2t 1
D2 3D
1þ
þ t
¼ e 6
2
4
1
1
1 D2 3D
¼ e2t t t
6
2
2 4
4
1
1
3
¼ e2t t þ :
6
2
8
The auxiliary equation is
4m2 þ 6m þ 1 ¼ 0;
pffiffiffiffiffiffiffiffiffiffi
pffiffi
pffiffi
which yields m ¼ 6 83616 ¼ 6 82 5 ¼ 3 4 5.
Therefore
pffi pffi 3þ 5
3 5
C.F ¼ c1 e 4 t þ c2 e 4 t ;
Further,
1
2
ð2et 6Þ ¼ et 6
4D2 þ 6D þ 1
11
2
¼ ð2x þ 3Þ 6:
11
Hence the solution is
P.I ¼
y ¼ C.F þ P.I
pffi
pffi
3þ 5
3 5
¼ c1 ð2x þ 3Þ 4 þ c2 ð2x þ 3Þ 4
2
þ ð2x þ 3Þ 6
11
EXAMPLE 1.199
Solve by using the method of variation of parameters ðD2 þ 9Þy ¼ cot 3x:
Ordinary Differential Equations
Solution. The auxiliary equation is m2 þ 9 ¼ 0;
which yields m ¼ 3i: Therefore,
C.F ¼ c1 cos 3x þ c2 sin 3x:
To find P. I, let FðxÞ ¼ cot 3x; y1 ¼ cos 3x: and
y2 ¼ sin 3x:
Then Wronskian
y y cos 3x
sin 3x ¼ 3:
W ¼ 01 02 ¼ y1 y2
3 sin 3x 3 cos 3x Therefore
P.I ¼ y1
Z
y2 FðxÞ
dx þ y2
W
Z
Z
y1 FðxÞ
dx
W
1
cos 3x sin 3x cot 3xdx
3
Z
1
þ sin 3x cos 3x cot 3xdx
3
Z
Z
1
1
cos2 3x
cos 3x cos 3xdx þ sin 3x
dx
¼
3
3
sin 3x
Z
1
1
1 sin2 3x
cos 3x sin 3x þ sin 3x
dx
¼
9
3
sin 3x
1
1
¼ cos 3x sin 3x þ sin 3x
9
3
Z
¼
ðcosec3x sin 3xÞdx
1
1
¼ cos 3x sin 3x þ sin 3x
9
3
1
cos 3x
logðcosec3x cot 3xÞ þ
3
3
1
¼ sin 3x½logðcosec3x cot 3xÞ
9
EXAMPLE 1.200
Solve ðD2 þ a2 Þy ¼ tan ax by the method of variation of parameters.
Solution. We have ðD2 þ a2 Þy ¼ tan ax. It’s auxiliary equation is m2 þ a2 ¼ 0, which yields
m ¼ ai:
Therefore
C. F ¼ c1 cos ax þ c2 sin ax:
To find particular integral, let
y1 ¼ cos ax and y2 ¼ sin ax:
n
Thus the wronskian W is given by
y
W ¼ 10
y1
Hence
cos ax
y2 ¼ y02 a sin ax
sin ax ¼ a:
a cos ax Z
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
P: I ¼ y1
W
W
Z
cos ax
¼
sin ax tan ax dx
a Z
sin ax
cos ax tan ax dx
þ
a
Z
cos ax
sin2 ax
dx
¼
a
cos ax
Z
sin ax
cos ax tan ax dx
þ
a
Z
cos ax
1 cos2 ax
dx
¼
a
cos ax
Z
sin ax
þ
cos ax tan ax dx
a
Z
cos ax
ðsec ax cos axÞdx
¼
a
Z
sin ax
þ
sin ax dx
a
cos ax
½log ðsec ax þ tan axÞ sin ax
¼
a2
1
2 sin ax cos ax
a
cos ax
¼
½log ðsec ax þ tan axÞ:
a2
Hence the complete solution of the given differential equation is
cos ax
y ¼ C.F þ P. I ¼ c1 cos ax þ c2 sin ax a2
½logðsec ax þ tan axÞ
EXAMPLE 1.201
Solve the differential equation by the method of variations of parameters:
y00 þ y ¼ sec2 x
Solution. The given differential equation is
Then
y01 ¼ a sin ax and y02 ¼ a cos ax:
1.105
y00 þ y ¼ sec2 x:
1.106
n
Engineering Mathematics-II
The symbolic form of the equation is
ðD2 þÞy ¼ sec2 x:
The auxiliary equation is m2 þ 1 ¼ 0, which yields
m ¼ i. Therefore
C.F ¼ c1 cos x þ c2 sin x:
To find particular integral, let
y1 ¼ cos x and
y2 ¼ sin x:
Then
y01 ¼ sin x and
y02 ¼ cos x:
The wronskian W is given by
y1 y2 cos x
¼
W ¼ 0
y1 y02 sin x
Therefore
sin x ¼ 1:
cos x Z
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
P.I ¼ y1
W
W
Z
¼ cos x sin x sec2 x dx
Z
þ sin x cos x sec2 x dx
Z
¼ cos x sec x tan x dx
Z
þ sin x sec x cos x dx
¼ cos x sec x þ sin x logðsec x þ tan xÞ
¼ 1 þ sin x logðsec x þ tan xÞ:
Hence the complete solution is
y ¼ C.F þ P.I
¼ c1 cos x þ c2 sin x 1 þ sin x logðsec x
þ tan xÞ:
EXAMPLE 1.202
Find the general solution of the equation
y00 þ 16y ¼ 32 sec 2x, using method of variation of
parameters.
Solution. The symbolic form of the given differential
equation is
ðD2 þ 16Þy ¼ 32 sec 2x:
The roots of the auxiliary equation m2 þ 16 ¼ 0 are
m ¼ ± 4i. Therefore the complementary function is
C.F ¼ c1 cos 4x þ c2 sin 4x:
To find the particular integral, let FðxÞ ¼ 32 sec 2x,
y1 ¼ cos 4x, y2 ¼ sin 4x. Then the Wronskian W is
given by
y y2 cos 4x
sin 4x ¼ 4:
W ¼ 10
y1 y02 4 sin 4x 4 cos 4x By the method of variation of parameters, we have
Z
Z
y2 FðxÞ
y1 FðxÞ
P.I ¼ y1
dx þ y2
dx
w
w
Z
sin 4xð32 sec 2xÞ
dx
¼ cos 4x
4
Z
cos 4xð32 sec 2xÞ
dx
þ sin 4x
4
Z
¼ 8 cos 4x sec 2xð2 sin 2x cos 2xÞ dx
Z
þ 8 sin 4x ð2cos2 2x 1Þ sec 2x dx
Z
¼ 16 cos 4x
sin 2x dx þ 8 sin 4x
Z
ðcos 2x sec 2xÞdx
cos 2x
þ 8 sin 4x
¼ 16 cos 4x
2
1
sin 2x log ðsec 2x þ tan 2xÞ
2
¼ 8 cos 4x cos 2x þ 8 sin 4x sin 2x
4 sin 4x log ðsec 2x þ tan 2xÞ
¼ 8 cos ð4x 2xÞ
4 sin 4x log ðsec 2x þ tan 2xÞ
¼ 8 cos 2x 4 sin 4x log ðsec 2x þ tan 2xÞ:
Hence the required solution is
y ¼ C.F þ P.I
¼ c1 cos 4x þ c2 sin 4x þ 8 cos 2x
4 sin 4x logðsec 2x þ tan 2xÞ
EXAMPLE 1.203
2
1
(a) Solve: ddx2y þ dy
dx 2y ¼ 1ex by using the method
of variation of parameters.
Ordinary Differential Equations
2
x
(b) Solve ddx2y þ 2 dy
log x by using the
dx þ y ¼ e
method of variation of parameters.
Solution.
(a) The symbolic form of the given differential
equation is
1
:
ðD þ D 2Þy ¼
1 ex
2
The corresponding auxiliary equation is
m2 þ m 2 ¼ 0;
which yields m ¼ 1 ; 2. Hence
C.F ¼ c1 ex þ c2 e2x :
Let
y1 ¼ ex ; y2 ¼ e2x
y02
2x
¼ 2e
Let FðxÞ ¼
so that
y01 ¼ ex
and
:
1
1ex .
Then Wronskian W is given by
y1 y2 e x
e2x ¼
W ¼ 0
y 1 y0 2 ex 2e2x ¼ 3ex e2x :
Therefore,
Z
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
W
W
Z
e2x
dx
¼ ex
x
2x
3e e ð1 ex Þ
Z
ex
þ e2x
dx
3ex e2x ð1 ex Þ
Z
Z
1
ex
1 2x
e2x
¼ ex
e
dx
dx:
1 ex
1 ex
3
3
P.I ¼ y1
But,
Z
ex
dx ¼
1 ex
Z
1
dz ; ex ¼ z
2
z ð1 zÞ
Z
1 1
1
þ 2þ
dz
¼
z z
1z
1
¼ log z logð1 zÞ
z
z
1
ex
ex
¼ log
¼ log
1 ex
1z z
and
Z
e2x
1
dx ¼
1 ex
2
Z
n
1.107
dz
1
¼ logð1 zÞ
1z
2
1
¼ logð1 ex Þ:
2
Therefore,
1
ex
1
ex þ e2x logð1 ex Þ:
P.I: ¼ ex log
x
1e
3
6
Hence the complete solution is
1
ex
y ¼ c1 ex þ c2 e2x þ ex log
ex
3
1 ex
1
þ e2x logð1 ex Þ:
6
(b) The auxiliary equation for the given differential
equation is
m2 þ 2m þ 1 ¼ 0;
Which yields m ¼ 1 ; 1. Thus
C.F ¼ ðc1 þ c2 xÞex :
To find P. I, let
y1 ¼ ex
Then,
x
e
W ¼ x
e
Therefore,
and
y2 ¼ x ex :
x ex
¼ e2x :
x ex þ ex Z
Z
y2 FðxÞ
y1 FðxÞ
dx þ y2
dx
W
W
Z
xex :ex log x
¼ ex
dx
e2x
Z x x
e :e log x
þ xex
dx
e2x
Z
Z
log x dx
¼ ex x log x dx þ x ex
P.I ¼ y1
¼ ex
x2
x2
log x þ x ex ½x log x x:
2
4
EXAMPLE 1.204
Solve: ðD2 3DD0 þ 2D02 Þz ¼ sin x cos y:
Solution. Replacing D by m and D0 by 1, the auxiliary
equation is
m2 3m þ 2 ¼ 0;
1.108
n
Engineering Mathematics-II
Which yields m ¼ 1; 2. Therefore
C: F ¼ 1 ðy þ xÞ þ 2 ðy þ 2xÞ:
Further,
1
sin x cos y
D2 3DD0 þ 2D02
1
¼ 2
D 3DD0 þ 2D02
1
½sin ðx þ yÞ þ sinðx yÞ
2
1
1
¼
sinðx þ yÞ
2 D2 3DD0 þ 2D02
1
sin ðx yÞ
þ 2
D 3DD0 þ 2D02
1
1
sin ðx þ yÞ
¼
0
2 ðD 2D Þ ðD D0 Þ
1
1
sin ðx yÞ
þ
2 ðD 2D0 Þ ðD D0 Þ
1
1
¼
2 ð1 2ÞðD D0 Þ
Z
cos u du; where u ¼ x þ y
P: I ¼
1
1
2 ðD 2D0 Þð1 þ 1Þ
Z
sin u du; where u ¼ x y
þ
1
sin ðx þ yÞ
2ðD D0 Þ
1
cos ðx yÞ
4ðD 2D0 Þ
1 x1
1 x1
sin
ðx
þ
yÞ
cos ðx yÞ
¼
2 12 :1!
4 1:1!
1
1
¼ x sin ðx þ yÞ x cosðx yÞ:
2
4
¼
EXAMPLE 1.205
A simple pendulum of length l is oscillating through
a small angle h in a medium in which the resistance
is proportional to velocity. Find the differential
equation of its motion. Discuss the motion and find
the period of oscillation.
Solution. Please refer to Figure 1.9 of Article 1.22.
We have jAOP ¼ h, AP ¼ s ¼ lh; OA ¼ l: The
resistance is proportional to the velocity and so it is
l ds
dt , where l is a constant. Therefore the equation of
motion along the tangent is
d2s
ds
m 2 ¼ mg sin h l :
dt
dt
or
d 2 ðlhÞ l dðlhÞ
þ g sin h ¼ 0
þ
dt2
m dt
or
d 2 h l dh g
þ sin h ¼ 0
þ
dt2 m dt l
Thus, to the first approximation of sinh; we have
d2h
dh gh
¼ 0;
ð168Þ
þ 2k þ
dt2
dt
l
where ml ¼ 2k. The differential equation (168)
describes the motion of the bob. Its auxiliary
equation is
g
m2 2km þ ¼ 0;
l pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
whose roots are m ¼ k
k 2 v2 ; where
g
2
v ¼ l : Since the oscillatory motion of the bob is
possible only if k < v; the roots are k
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
i v2 k 2 : Therefore the solution of differential
equation (168) is
h
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
h ¼ ekt C1 cos
v2 k 2 t
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi i
þC2 sin
v2 k 2 t ;
2
which gives a vibratory motion of period pffiffiffiffiffiffiffiffiffiffi
.
v2 k 2
EXAMPLE 1.206
(a) Solve the simultaneous equations dx
dt y ¼ t and
dy
2
þ
x
¼
t
.
dt
(b) Solve the following simultaneous equations
dy
dx
dt ¼ 3x þ 2y and dt ¼ 5x þ 3y.
(c) Solve the following simultaneous equations
dy
dx
dt þ 5x 2y ¼ t, dt þ 2x þ y ¼ 0.
Solution. (a) We have
Dx y t ¼ 0;
x þ Dy t ¼ 0:
2
ð169Þ
ð170Þ
Multiplying (169) by D and adding to (170), we get
D2 x Dt t2 ¼ 0
Ordinary Differential Equations
or
2
or
D2 x
¼ 1 þ t2 :
dt2
Therefore
dx
t3
¼ t þ þ c1
dt
3
Hence
and
t2 t4
x ¼ þ þ c1 x þ c2 :
2 12
Putting the value of dx
dt in (169), we get
t3
t ¼ þ c1 y t ¼ 0
3
or
t3
y ¼ þ c1 :
3
Thus the required solution is
t2 t4
x ¼ þ þ c1 x þ c2 ;
2 12
t3
y ¼ þ c1 :
3
(b) The given equations are
dx
dy
¼ 3x þ 2y and
¼ 5x þ 3y:
dt
dt
In symbolic form, we have
ðD 3Þx 2y ¼ 0;
ð171Þ
5x þ ðD 3Þy ¼ 0
ðD 3Þ2 x 2ðD 3Þy ¼ 0
ð172Þ
10x þ 2ðD 3Þy ¼ 0
ð173Þ
2x þ ðD þ 1Þy ¼ 0
ð177Þ
Operating (176) by ðD þ 1Þ and multiplying (177)
by 2 and then adding, we get
ðD2 þ 6D þ 9Þx ¼ 1 þ t:
The auxiliary equation is
m2 þ 6D þ 9 ¼ 0
and so m ¼ 3; 3 and hence
C.F ¼ ðc1 þ c2 tÞe3t
Further,
1
D 2
1þ
ð1 þ tÞ ¼
ð1 þ tÞ
P.I ¼
9
3
ðD þ 3Þ2
1
2D
1
1
þ ð1 þ tÞ ¼
tþ :
¼ 1
9
3
9
3
1
Hence,
3t
1
1
t
þ
ð178Þ
þ
9
3
ð174Þ
Therefore,
dx
1
¼ 3ðc1 þ c2 tÞe3t þ c2 e3t þ :
dt
9
Putting in the first (given) equation, we get
1
2
4
y ¼ ðc1 þ c2 tÞe3t þ c2 e3t t þ
ð179Þ
2
9
27
Hence the solution is given by (178) and (179).
ð175Þ
EXAMPLE 1.207
Solve the following simultaneous equations
dx
dy
þ 2x þ 3y ¼ 0; 3x þ þ 2y ¼ 2et
dt
dt
Adding (172) and (173), we get
½ðD 3Þ2 x 10x ¼ 0
or
The auxiliary equation of (174) is
m2 6m 1 ¼ 0;
pffiffiffiffiffi
which gives m ¼ 3
10. Therefore
pffiffiffiffiffi
pffiffiffiffiffi
3t
x ¼ e ½c1 cosh 10t þ c2 sin 10t
pffiffiffiffiffi
pffiffiffiffiffi
1 pffiffiffiffiffi
y ¼ ½e3t 10½c1 sinh 10t þ c2 cosh 10t:
2
(c) The symbolic forms of the equations are
ðD þ 5Þx 2y ¼ t
ð176Þ
x ¼ C.F þ P.I ¼ ðc1 þ c2 tÞe
or
ðD2 6D 1Þx ¼ 0
1.109
From equation (171), we have (using (175))
dx
2y ¼ 3x
dt pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
¼ e3t 10½c1 sinh 10t þ c2 cosh 10t
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
þ 3 e3t 10½c1 cosh 10t þ c2 sinh 10t
pffiffiffiffiffi
pffiffiffiffiffi
pffiffiffiffiffi
3 e3t 10½c1 cosh 10t þ c2 sinh 10t
pffiffiffiffiffi
pffiffiffiffiffi
¼ e3t ½c1 sinh 10t þ c2 cosh 10t:
D x1t ¼0
2
n
1.110
n
Engineering Mathematics-II
Solution. In symbolic representation, we have
ðD þ 2Þx þ 3y ¼ 0
ð180Þ
and
3x þ ðD þ 2Þy ¼ 2et :
ð181Þ
2
ð2nÞ!
.
(d) P2n ð0Þ ¼ ð1Þ
22n ðn!Þ2
Solution.
(a) We have (see solved Exercise 45, chapter 1)
d
½ð1 x2 ÞP0n ðxÞPn ðxÞ þ nðn þ 1ÞPn ðxÞ ¼ 0
dx
Multiplying (180) by ðD þ 2Þ and (181) by 3 and
subtracting, we get
ðD þ 2Þ2 x 9x ¼ 6et
or
or
ðD2 þ 4D 5Þx ¼ 6et
nðn þ 1ÞPn ðxÞ
d
¼ ½ð1 x2 ÞP0n ðxÞPn ðxÞ
dx
ð182Þ
The auxiliary equation for (182) is m2 þ 4m 5 ¼
0; whose roots are 1 and 5. Therefore
C.F ¼ c1 et þ c2 e5t :
Further,
Multiplying both sides of (185) by Pn ðxÞ and integrating between the limits –1 and 1, we get
Z1
1
ð6et Þ
2
D þ 4D 5
1
t
ð6et Þ ¼ ð6et Þ
¼t
2D þ 4
6
¼ tet :
P.I ¼
Hence the complete solution of (182) is
x ¼ c1 et þ c2 e5t tet
ð185Þ
nðn þ 1Þ
P2n ðxÞdx
1
Z1
¼
½ð1 x2 ÞP0n ðxÞ0 Pn ðxÞdx
1
¼ fPn ðxÞ½ð1 x2 ÞP0n ðxÞg11
ð183Þ
Differentiating (183) with respect to t, we have
dx
¼ c1 et 5c2 e5t ½et þ tet :
dt
dx
Putting the value of dx
dt in the equation dt þ 2x þ
3y ¼ 0; we get
dx
3y ¼ 2x ¼ c1 et 5c2 e5t et tet
dt
2½c1 et þ c2 e5t tet Z1
þ
1
Z1
¼
ð1 x2 Þ½P0n ðxÞ2 dx
1
But
¼ 3c1 et þ 4c2 e5t þ et þ 3tet
Z1
P2n ðxÞdx ¼
or
4
1
ð184Þ
y ¼ c2 e5t c1 et þ et þ tet
3
3
Expressions (183) and (184) represent the required
solution.
EXAMPLE.1.208 1
R
(a) Show that ð1 x2 Þ½P0 n ðxÞ2 dx ¼ 2nðnþ1Þ
2nþ1 :
1R1
1
n R
(b) Show that
f ðxÞPn ðxÞdx ¼ ð1Þ
ðx2 1Þn
2n n!
ðnÞ
1
1
f ðxÞdx.
(c) Show that J3 ðxÞ þ 3J00 ðxÞ þ 4J0000 ðxÞ ¼ 0:
P0n ðxÞð1 x2 ÞP0n ðxÞdx
1
2
:
2n þ 1
Therefore (186) implies
Z1
ð1 x2 ÞðP0n ðxÞ2 dx ¼
1
2nðn þ 1Þ
:
2x þ 1
(b) By Rodrigue’s formula, we have
Pn ðxÞ ¼
1
dn
2n n ! dxn
ðx2 1Þn :
ð186Þ
Ordinary Differential Equations
Therefore integrating by parts , we have
Z1
1
1
f ðxÞPn ðxÞdx ¼ n
2 n!
1
¼ n
2 n!
Z1
"
Z1
f 0 ðxÞ
1
ð1Þ
¼ n
2 n!
Z1
1
ð1Þ2
¼ n
2 n!
d n1 2
n
ðx 1Þ
dxn1
1
3
d n1 2
n
ðx 1Þ dx5
dxn1
d n1
f ðxÞ n1 ðx2 1Þdx
dx
4J0000 ðxÞ þ 3J00 ðxÞ þ J3 ðxÞ ¼ 0:
(d) Since ð1 2xt þ t2 Þ2 is generating function of
the Legendre’s polynomial, we have
1
Z1
f ðxÞ
1
d n1 2
ðx 1Þn dx
dxn1
::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::::
::::::::::::::::::::::::::::::::::
Z1
Hence
0
(using integration by parts second time)
ð1Þn
¼ n
2 n!
1
J0000 ðxÞ ¼ ½J00 ðxÞ J20 ðxÞ
2
1
1 1
¼ J00 ðxÞ þ
fJ1 ðxÞ J0 ðxÞg
2
2 2
1
1
1
¼ J00 ðxÞ þ J1 ðxÞ J3 ðxÞ
2
4
4
1 0
1
1
¼ J0 ðxÞ þ ½J00 ðxÞ J3 ðxÞ
2
4
4
3 0
1
¼ J0 ðxÞ J3 ðxÞ:
4
4
1
f ðxÞ
1.111
and
dn
f ðxÞ n ðx2 1Þn dx
dx
1
n
f ðnÞ ðxÞðx2 1Þn dx
1
(using integration by parts nth time)
(c) We know that
d n
½x Jn ðxÞ ¼ xn Jnþ1 ðxÞ
dx
(Please see Example 1.151 (ii))
Putting n ¼ 0, we have
J00 ðxÞ ¼ J1 ðxÞ:
Therefore
J000 ðxÞ ¼ J10 ðxÞ
1
¼ ½J0 J2 2
(Please see Example 1.151 (iv))
ð1 2x þ t2 Þ2 ¼ tn Pn ðxÞ
1
or
X
1
1:3 4
tn Pn ð0Þ ¼ 1 t2 þ
t
2
2:4
1:3:5 ð2r 1Þ 2r
t
ð1Þr
2:4:6 ð2rÞ
þ Equating coefficients of t2n on both sides, we get
1:3:5 ð2n 1Þ
2:4:6 ð2nÞ
1:2:3:4 ð2n 1Þ2n
¼ ð1Þn
½2:4:6 ð2nÞ2
ð2nÞ !
ð2nÞ !
¼ ð1Þn
¼ ð1Þn
:
2
n
2n
½2 ð1:2 nÞ
2 ðn !Þ2
P2n ð0Þ ¼ ð1Þn
EXAMPLE 1.209
Solve the simultaneous equations
dx
dy
þ y ¼ sin t;
þ x ¼ cos t;
dt
dt
given that x ¼ 2, y ¼ 0 when t ¼ 0
Solution. We have
dx
þ y ¼ sin t
dy
and
dy
þ x ¼ cos t;
dt
1.112
n
Engineering Mathematics-II
under the condition that x ¼ 2, y ¼ 0 when t ¼ 0. In
symbolic form, we have
Dx þ y ¼ sin t
ð187Þ
Dy þ x ¼ cos t:
ð188Þ
which are analytic at x ¼ 0. Hence x ¼ 0 is a regular
singular point. So, let
1
X
an xnþm ; an 6¼ 0:
y¼
Multiplying (187) by D and subtracting (188) from
the resultant equation, we have
D2 x x ¼ cos t cos t ¼ 0:
Differentiating twice, we get
1
dy X
ðn þ mÞan xnþm1
¼
dx n¼0
Its Auxiliary equation is
m2 1 ¼ 0; which yields
m¼
n¼0
1
d2y X
¼
ðn þ mÞðn þ m 1Þan xnþm2 :
dx2 n¼0
1:
Hence
x ¼ c1 et þ c2 et :
When t ¼ 0, we have x ¼ 2. Therefore
c 1 þ c2 ¼ 2
ð189Þ
ð190Þ
Now putting the value of x from (189) in (187), we
get
Dðc1 et þ c2 et Þ þ y ¼ sin t
2
d y
Substituting the values of y, dy
dx and dx2 in the given
equation, we have
1
1
X
X
2ðn þ mÞðn þ m 1Þan xnþm ðn þ mÞan xnþm
n¼0
n¼0
þ
1
X
an xnþmþ1 n¼0
or
or
1
X
c1 et c2 et þ y ¼ sin t:
or
y ¼ c1 e þ c2 e
t
t
þ sin t
When t ¼ 0, y ¼ 0. Therefore
0 ¼ c1 þ c2
ð191Þ
5an xnþm ¼ 0
n¼0
½2ðn þ mÞðn þ m 1Þ ðn þ mÞ 5an xnþm
n¼0
þ
ð192Þ
Solving (190) and (192), we have c1 ¼ 1 ; c2 ¼ 1.
Hence the required solution is
x ¼ et þ et ; y ¼ et þ et þ sin t:
1
X
1
X
an xnþmþ1 ¼ 0
n¼0
or
1
X
½2n2 þ 2m2 þ 4mn 3n 3m 5an xnþm
n¼0
EXAMPLE 1.210
Solve the following differential equation in series:
d2y
dy
2x2 2 x þ ðx 5Þy ¼ 0:
dx
dx
Solution. Comparing the given equation with
d2y
dy
þ PðxÞ þ QðxÞy ¼ 0;
d x2
dx
we get
1
x5
PðxÞ ¼ ; QðxÞ ¼
:
2x
2x2
At x ¼ 0, both PðxÞ and QðxÞ are not analytic. Thus
x ¼ 0 is a singular point. Further
1
x5
and x2 QðxÞ ¼
;
xPðxÞ ¼ 2
2
þ
1
X
an xnþmþ1 ¼ 0
n¼0
or
1
X
½2n2 þ 2m2 þ 4mn 3n 3m 5an xnþm
n¼0
þ
1
X
an xnþmþ1 ¼ 0
n¼0
or
1
X
½2n2 þ 2m2 þ 4mn 3n 3m 5an xnþm
n¼0
þ
1
X
n¼1
an1 xnþm ¼ 0
Ordinary Differential Equations
or
n
1.113
Therefore the general solution is
ð2m2 3m 5Þa0 xm þ
1
X
y ¼ c 1 y1 þ c2 y2
½ð2n2 þ 2m2 þ 4mn
x
x2
x3
5
þ ¼ Ax2 1 þ þ
9 198 7722
x x2 x3
þ Bx1 1 þ þ þ þ :
5 30 90
n¼1
3n 3m 5Þan þ an1 x
mþn
¼ 0:
Therefore the indicial equation is
5
; 1:
2
Equating to zero the other coefficients, we have
2m2 3m 5 ¼ 0;
m¼
which yields
ð2n þ 2m þ 4mn 3n 3m 5Þan
2
2
ð193Þ
¼ an1
for
n
1:
For m ¼ 52, we have from (193),
ð2n2 þ 7nÞan ¼ an1 ; a
1
or
an1
; n 1:
2n2 þ 7n
Thus, putting n ¼ 1 ; 2 ; 3; we get
a0
a1 ¼ 9
a0
a2 ¼
198
a0
a3 ¼ 7722
and so on. Therefore
x
x2
x3
5
þ :
y 1 ¼ a0 x 2 1 þ
9 198 7722
EXAMPLE 1.211
For Bessel’s polynomial Jn ðxÞ of degree n, show
that
(a) J20 ðxÞ ¼ 1 x42 J1 ðxÞ þ 2x J0 ðxÞ
(b) J02 þ 2 J12 þ J22 þ J32 þ ¼ 1.
Solution.
(a) We know that
1
Jn0 ðxÞ ¼ ½Jn1 ðxÞ Jnþ1 ðxÞ:
2
an ¼ Therefore
1
J20 ðxÞ ¼ ½J1 ðxÞ J3 ðxÞ:
2
Also
Jnþ1 ðxÞ ¼
or
an ¼
an1
; n
2n2 7n
4
J3 ðxÞ ¼ J2 ðxÞ J1 ðxÞ
x
4 2
¼
J1 ðxÞ J0 ðxÞ J1 ðxÞ
x x
8
4
¼ 2 1 J1 ðxÞ J0 ðxÞ:
x
x
1
1:
Therefore,
a0
a0
¼
5
5
a1
a0
¼
a2 ¼ 6 30
a2
a0
¼
;
a3 ¼ 3 90
and so on. Hence
x x2 x3
y2 ¼ a0 x1 1 þ þ þ þ 5 30 90
a1 ¼ 2n
Jn ðxÞ Jn1 ðxÞ:
x
Therefore
For m ¼ 1, the relation (1) yields
ð2n2 7nÞan ¼ an1 ; n
ð194Þ
Putting this value in (194), we get
4
2
0
J2 ðxÞ ¼ 1 2 J1 ðxÞ þ J0 ðxÞ:
x
x
(b) From example 1.151, the Jacobi’s series are
J0 þ 2J2 cos 2 þ 2J4 cos 4 þ ¼ cosðx sin Þ
and
2J1 sin þ 2J3 sin 3 þ 2J5 sin 5 þ ¼ sinðx sin Þ:
1.114
n
Engineering Mathematics-II
Squaring these equations and integrating with
respect to in the interval ð0; Þ, we get
J02 þ 2 J22 þ 2J 24 þ Z
¼ cos2 ðx sin Þd
ð195Þ
0
ð196Þ
0
Adding (195) and (196) we get
Z
J02
þ 2
J12
þ
2 J22
þ ¼
dy
dx ¼
y0 ¼ yþx
x
(c) 2xy
and
2J12 þ 2 J32 þ 2 J52 þ Z
¼ sin2 ðx sin Þd
(b) ydx xdy ¼
d ¼ :
0
J02 þ 2 J12 þ 2 J22 þ 2J32 þ ¼ 1:
3y2 þx2
Ans. x2 + y2 ¼ cx3
(d)
Ans. y ¼ x log |cx|
4. Reduce the following equations to homogeneous form and solve them
(a) (2x + y 3)dy ¼ (x + 2y – 3)dx
Ans. (y – x)3 ¼ k(y + x – 2)
(b) (x + 2y) (dx – dy) ¼ dx + dy
Ans. logðxþyþ 13Þ þ 32 ðy xÞ þ k
axþhyþg
þ
(c) dy
dx
hxþbyþf ¼ 0:
Ans. ax2 + 2hxy + by2 + 2gx + 2fy + c ¼ 0
5. Solve the following linear equations :
2
(a) x2 dy
dx ¼ 3x 2yx þ 1
Ans. y ¼ xþ 1x þ xc2 :
(b) ðxþ1Þ
Hence
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
x2 þ y2 dx
pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ans. y þ x2 þ y2 ¼ c
(c)
dy
dx
dy
dx
¼ y ¼ e3x ðx þ 1Þ2 :
Ans. y¼ ðxþ1Þð13 e3x þ cÞ
xþy cos x
1þsin x
2
EXERCISES
1. Form differential equation from the following
equations
(a) y ¼ a e2x þ be3x þ c e x
2
Ans. x ddx2y þ 2 dy
dx xy ¼ 0
2
(b) x ¼ A cos (nt + a) Ans. ddt2x þ n2 x ¼ 0
2. Solve the separable equations
(a) (x 4) y4 dx x3(y3 3) dy ¼ 0.
Ans. 1x þ x22 þ 1y y13 ¼ c
¼ e2x3y þ 4x2 e3y
Ans. 3e x 2e3y þ 8x3 ¼ c
dy
(c) dx x tanðy xÞ ¼ 1:
Ans. log sin (y x) ¼ 12x2 + c
(b)
dy
dx
2
(d) ðx yÞ2 dy
dx ¼ a :
Ans. alog
(e)
dx
dt
xya
xyþa
¼ 2y þ k
¼ x 2xþ2
2
Ans.pxffiffiffi ¼ 1 + tan (t + c)
þ
cot
y¼
0
if
yð 2Þ ¼ =4:
(f) x dy
dx
Ans. x¼ 2 cos y
3. Solve the homogeneous equations :
(a)
dy
dx
¼ yx þ sin yx
Ans. 2x tan1 ðcxÞ
Ans. yð1 þ sin xÞ ¼ c x2
(d) y log y dx + (x – log y) dy ¼ 0
Ans. x ¼ 12 log y þ logc y
2
c
(e) xlog x dy
dx þy¼ log x : Ans. y¼ log xþ log x
6. Solve the following equations
(a) (x3 y2 + xy) dx ¼ dy.
2
Ans. 1y ¼ x2 2 þ cex =2
(b)
dy
dx
y
x
tan
1þx ¼ ð1 þ xÞe sec y
Ans. sin y ¼ (1 + x) (ex + c)
þ xz log z ¼ xz ðlog zÞ2
Ans. log1 z ¼ 1 þ cx
dy
3
(d) dx þ y tan x ¼ y sec x
Ans. cos2 x ¼ y2(c + 2 sin x)
7. Solve the exact equations
(a) [y (1 + 1x) + cos y]dx + (x + log x – x sin y)dy ¼ 0
Ans. y(x + log x) + x cos y ¼ 0
(c)
dz
dx
(b) [5x4 + 3x2 y2 2xy3]dx ¼ (2x3y – 3x2 y2 – 5y4)dy ¼ 0
Ans. x5 + x3 y2 – x2 y3 – y5 ¼ 0
(c)
2x
y3
dx þ y
2
3x2
y4
dy ¼ 0
Ans. x2 – y2 ¼ cy3
(d) (x + y – a ) x dx + (x2 – y2 –b2)y dy ¼ 0.
2
2
2
Ans. x4 + 2x2 y2 – y4 – 2a2x2 – 2b2y2 ¼ c
Ordinary Differential Equations
8. Solve the following equations which are reducible to exact equations:
(a) (y2 + 2x2y) dx + (2x3 xy)dy ¼ 0.
pffiffiffiffiffi 3=2
¼ c
Ans. 6 xy yx
2
2
(b) (x + y + 1) dx 2xy dy ¼ 0.
2
Ans. x2 1x yx ¼ c
(c) (xy ex/y + y2)dx x2 ex/y dy ¼ 0
Ans. 3log x 2 log y +yx¼ c
(d) x dy – y dx + a(x2 + y2)dx ¼ 0
Ans. axþ tan1 yx ¼ c
2
2
Ans. x2 þ xy ¼ c
(e) x dy – y dx ¼ xy dx
3
x3 þy3
(f) dy
Ans. 3 log x yx ¼ c
dx ¼ xy2
9. In RC circuit containing steady voltage V,
find the change at any time t. Also derive
expression for the current flowing through the
circuit.
V
t
t
Ans Q ¼ CV ð1 e RC and I ¼ e RC
R
:
10. An RL circuit has an e.m.f. of 3 sin 2t volts, a
resistance of 10 ohms, an inductance of 0.5
henry, and an initial current of 6 amp. Find the
current in the circuit at any time t.
:
30
3
20t
þ 101
sin 2t 101
cos 2t
Ans I ¼ 609
101 e
11. If the air is maintained at 30 C and the temperature of the body cools from 80 C to 60 C in
12 min, find the temperature of the body after
24 min.
Ans. 48 C
12. A pipe 20 cm in diameter contains steam at
200 C. It is covered by a layer of insulation 6
cm thick and thermal conductivity 0.0003. If
the temperature of the outer surface is 30 C,
find the heat loss per hour from 2 metre length
of the pipe.
Ans. 490,000 cal.
13. A steam pipe 20 cm in diameter is covered with
an insulating sheath 5 cm thick, the conductivity
of which is 0.00018. If the pipe has the constant
temperature 100 C, and outer surface of the
sheath is kept at 30 C, find the temperature of
the sheath as a function of the distance x from
the axis of the pipe. How much heat is lost per
hour through a section 1 metre long.
Ans. T ¼ 497.5 – 172.6 log x and
Q ¼ 70300 cal.
n
1.115
14. A tank contains 5000 litres of fresh water. Salt
water which contains 100 gm of salt per litre
flows into it at the rate of 10 l/min and the
mixture kept uniform by stirring, runs out at the
same rate. When will the tank contain 2,00,000
gm of salt? How long will it take for the
quantity of salt in the tank to increase from
1,50,000 gm to 2,50,000 gm?
Ans. 4 h 15.52 min, 2 h 48.23 min
15. The amount x of a substance present in a certain
chemical reaction at time t is given by dx
dt þ
x
10t
¼
2
1:5
e
:
If
at
t
¼
0,
x
¼
0.
5,
find
x at
10
69
t ¼ 10.
Ans 20 2e
:
16. Radium decomposes at a rate proportional to
the amount present. If a fraction p of the original amount disappears in 1 year, how much
will it remain at the end of 21 years?
Hint: use x ¼ x0 ekt. Using given data k ¼ log 1p]
Ans. ð1 1pÞ21 times the original amount
17. Find the orthogonal trajectories of the family of
curves r ¼ a(1+cos h).
Ans. r ¼ c(1 – cos h)
18. Find the orthogonal trajectories of the family of
parabolas y2 ¼ 4ax.
Ans. 2x2 + y2 ¼ c
19. Find the orthogonal trajectories of the family of
curves ay2 ¼ x3 .
Ans. 3y2 + 2x2 ¼ c2
20. Solve the following differential equation
(a)
d4 y
dx4
3
2
4 ddx3y þ 14 ddx2y 20 dy
dx þ 25y ¼ 0
Hint: Roots of A.E. are 1 + 2i, 1 – 2i, 1 + 2i, 1 – 2i.
Ans. y ¼ ex[(c1 + c2x) sin 2x + (c3 + c4x)cos 2x
2
0
(b) ddx2y 6 dy
dx þ25y ¼ 0; y(0) ¼ –3, y (0) ¼ –1.
Ans. y ¼ e3x(2 sin 4x 3 cos 4x)
(c)
d4 y
dx4
2
þ 8 ddx2y þ 16y ¼ 0:
Ans. y ¼ (c1 + c2x) cos 2x + (c3 + c4x) sin 2x
(d)
(e)
d2 x
dt2
2
d y
dx2
þ 6 dx
dt þ 9x ¼ 0
þða þ bÞ
dy
dx
Ans. x ¼ (c1 + c2t) e–3t
þ aby ¼ 0
Ans. y ¼ c1 eax þ c2 ebx
(f) ðD þ D þ 4D þ 4Þ y ¼ 0
3
2
1.116
n
Engineering Mathematics-II
Ans. y ¼ c1 e + c2 cos 2x + c3 sin2x
–x
d3 y
dx2
2 dy
dx þ 2y ¼ 0: Are the solutions linearly independent?
Ans. y ¼ c1 e x þ c2 e x ; Wronskian
e x e2x ¼ e3x 6¼ 0;
Wðe x ; e2x Þ ¼ x
e 2e2x Therefore, ex,e2x are linearly independent.
(g)
21. Solve the following differential equations:
2
(a) ddx2y þ a2 y ¼ tan ax
Ans. y ¼ c1 cos ax + c2 sin ax
– a12 cos ax log (sec ax + tan ax)
(b) (D2 3D + 2) y ¼ 6 e–3x + sin 2x.
:
:
Ans y ¼ c1 þðc2 þc3 xÞex þ e18 ðx2 78 xþ 11
6Þ
1
þ
ð3 sin 2xþ4 cos 2xÞ
100
(k)
d2y
dx2
d2 y
dx2
4y ¼ x2 þ 2x:
Ans. y ¼ c1 e2x þ c2 e2x 14 ðx2 þ 12Þ
(d) (D2 4D + 4)y ¼ 8x2 e2x sin 2x.
Ans y ¼ ðc1 þ c2 xÞe2x
:
e2x ½4x cos 2x þ ð2x2 3Þ sin 2x
(e) (D 1)y ¼ x sin x + (1 + x2)ex.
1
Ans y ¼ c1 e x þc2 ex ðx sin x þ cos xÞ
2
1
x
2
þ
xe ð2x 3xþ9Þ
12
( f ) (D 1)2 (D + 1)2 y ¼ sin2 2x + ex + x.
2
:
þ 12 18 cos xþe x þx
(g) (D2 + 4)y ¼ ex + sin 2x.
Ans. y ¼ c1 cos 2xþc2 sin 2xþ 15 e x 4x cos 2x
(i)
d3 y
dx2
3
2
þ 3 dy
dx þ 2y ¼ 4 cos x :
Ans y ¼ c1 ex þ c2 e2x þ 1
:
d y
dx2
3 dy
dx
:
þ
1
10
3x
Ans. y ¼ c1 + c2 ex + c3 e–x
+ x ex (x2 + x) – 2 sin x
22. Solve the following differential equations of
second order by changing the independent
variable.
2
(a) x4 ddxy2 þ 2x3 dy
dx þ 4y ¼ 0
ð3 sin 2x cos 2xÞ
þ 2y ¼ xe þ sin 2x :
d3 y
dx3
1
þ 20
ð3 cos 2x sin 2xÞ
2
2
2 2x
þ 2 ddx2y þ dy
dx ¼ x e þ sin x :
2
x
2
2
1
(b) x6 ddx2y þ 3x5 dy
dx þ a y ¼ x2
Ans. y ¼ c1 cos 2xa2 c2 sin 2xa2 þ a21x2
2
(c) x2 ddxy2 x dy
dx þ y ¼ log x
Ans. y ¼ ðc1 þ c2 log xÞx þ log x þ 2
(d)
2
cos x ddxy2
3
5
þ sin x dy
dx ð2cos xÞy ¼ 2cos x
pffiffi
pffiffi
Ans. y ¼ c1 e 2 sin x þ c2 e 2 sin x þ sin2 x:
23. Solve the following differential equations of
second order by changing the dependent
variable:
2
2
2
(a) ddx2y þ 2n cot nx dy
dx þ ðm n Þy ¼ 0
Ans. y ¼ sin1nx ½c1 cos mx þ c2 sin mx:
2
(b) x2 ddx2y 2y ¼ x2 þ 1x
Ans. y ¼ c1 x2 þ cx2 þ 13 x2 1x log x:
2
2
(c) cos2 x ddxy2 ð2 sin x cos xÞ dy
dx þ ðcos xÞy ¼ 0
pffiffiffi
pffiffiffi
Ans.
2xÞ sec x:
h 2 y ¼i ½c1 cosð 2xÞ þ c2 sinð
d y
dy
(d) dx2 þ y cot x þ 2 y tan x þ dx ¼ 0
Ans. y ¼ ðc1 þ c2 xÞ cos x:
24. Solve by method of undetermined coefficients:
2
(a) ddx2y þ y ¼ 2ex þ cos x
Ans. y ¼ c1 cos x þ c2 sin x þ ex þ 12 x sin x
2
2 x
(b) ddx2y 2 dy
dx þ y ¼ x e
Ans y ¼ c1 e x þ c2 e 2x þ 14 e3x ð2x 3Þ
( j)
x
(l) (D2 D)y ¼ 2x + 1 + 4 cos x + 2ex.
Ans y ¼ ðc1 þ c2 xÞe x þ ðc3 þ c4 xÞex
(h)
:
Ans. y ¼ c1 cos 2x þ c2 sin
3 3x
1
þ 10
e þ 20
ð3 cos 2x sin 2xÞ
:
x
e
þ 3 dy
dx þ2y ¼ e :
Ans y ¼ c1 ex þc2 e2x þe2x ee
Ans y ¼ c1 e x þ c2 e2x
(c)
2x
4
x x
Ans. y ¼ ðc1 þ c2 xÞex þ 12
e
2
x
(c) ddx2y þ 2 dy
dx þ y ¼ x e
Ans. y ¼ ðc1 þ c2 xÞex þ x 2 e4 :
x
Ordinary Differential Equations
25. Solve by method of reduction of order.
3
n
1.117
2
(d) x2 ddx3y 4x ddxy2 þ 6 dy
dx ¼ 4
2
3
(a) x2 ddxy2 2xð1 þ xÞ dy
dx þ 2ð1 þ xÞy ¼ x
if y ¼ x is a solution of the C. F.
Hint: Write the given equation in the form
d3y
d2y
dy
x3 3 4x2 2 þ 6x
¼ 4x
dx
dx
dx
Ans. y ¼ x 2x þ c21 e2x þ c2 :
3
2
2 2
(b) ð1 x2 Þ ddx2y þ x dy
dx y ¼ xð1 x Þ
if y ¼ x is a parth of C. F.
i
3
pffiffiffiffiffiffiffiffiffiffiffiffiffi
2 2
Þ
2 þ xsin1 x þ c x:
1
x
c
Ans. y ¼ xð1x
1
2
9
26. Solve the following equations using the method
of variation
of parameters:
2
(a) ddxy2 þ a2 y ¼ sec ax:
:
Ans y ¼ c1 cos ax þ c2 sin ax
þ a12 cos ax logðcos axÞþ 1a x sin ax
(b)
d2y
dx2
þy ¼ x sin x:
d2y
dx2
+ 4y ¼ tan 2x.
:
x
x2
Ans y ¼ c1 cos x þ c2 sin xþ sin x cos x
4
4
(c)
:
(d)
d2y
dx2
1
4
x
Ans. y ¼ 1x ðc1 þ c2 log xÞ þ 1x log 1x
2
sinðlogxÞþ1
(f) x2 ddx2y 3x dy
dx þ y ¼ log
x
pffiffi
pffiffi
3þ2
2 3
Ans y ¼ c1 x
þ c2 x
:
1
þ 61x
log x½5sinðlog xþ6cosðlog xÞ
þ
cos 2 x logðsec 2x þ tan 2xÞ
x
2 dy
dx þ 2y ¼ e tan x :
Ans. y ¼ ex(c1 cos x + c2 sin x)
ex cos x log (sec x + tan x)
1
(e) ðD2 þ 1Þy ¼ 1þsin
x :
Ans. y ¼ c1 cos x + c2 sin x
+ sin x log(1 + sin x) – x cos x – 1
2
(f ) (D + 1)y ¼ x sin x.
2
Ans y ¼ c1 cos x þ c2 sin xþ 4x sin x x4 cos x
27. Solve the following Cauchy–Euler equations :
3
2
(a) x3 ddx3y þ 2x2 ddx2y þ 2y ¼ 10 x þ 1x :
c1
Ans y ¼ þx½c2 cosðlogxÞþc3 sinðlogxÞ
x
:
2
2
(b) x2 ddx2y 3x dy
dx þ 5y ¼ x sinðlog xÞ
Ans y ¼ x ½c1 cos ðlog xÞþ c2 sin ðlog xÞ
(c)
d2 y
dx2
2
12 x2 log x cos ðlog xÞ
2
x2 2x dy
dx 4y ¼ x þ 2 log x
6
Ans y ¼ cx1 þ c2 x4 x6 12 log x
:
½21sinðlog xÞ þ 191cosðlog xÞ
1
ð1 þ log xÞ
6x 2
3
d
y
d
(g) x3 dx3 þ 3x2 dx2y þ x dy
dx þ 8y ¼ 65 cosðlog xÞ
pffiffiffi
2
Ans y ¼ c1 x þ x½c2 cos 3ðlog xÞ
pffiffiffi
þ c3 sinð 3 log xÞ
þ 8cosðlog xÞ sinðlog xÞ
þ
28. Solve the following Legendre’s linear
equations:
2
2
(a) ð3xþ2Þ ddx2y þ3ð3xþ2Þ dy
dx 36y¼ 3x þ4xþ1:
:
Ans y ¼ c1 ð3x þ 2Þ2 þ c2 ð3xþ2Þ2
1
þ 108
½ð3xþ2Þ2 logð3xþ2Þ þ 1
(b) ðxþ1Þ2
d2y
dx2
þ ðx þ 1Þ
dy
dx
¼ ð2xþ3Þð2xþ4Þ:
Ans. y ¼ c1 + c2 log (x + 1)
+ [log(x + 1)]2 + x2 + 8x
:
:
2
61
:
Ans y ¼ c1 cos 2x þ c2 sin 2x
Ans. y ¼ c1 þ c2 x3 þ c3 x4 þ 23 x
2
(e) x2 ddx2y 4x dy
dx þ 6y ¼ x
2
2
2
(c) ð1þ2xÞ2 ddx2y 6ð1þ2xÞ dy
dx þ16y ¼ 8ð1þ2xÞ
Ans. (1 + 2x)2 [c1 + c2 log (1 + 2x) + log(1 + 2x)]
29. Solve the following simultaneous equations:
(a)
d2 x
dt2
:
þ 4x þ 5y ¼ t2 ;
Ans x ¼ c1 e þ c2 e
t
t
d2 y
dt2
þ 5x þ 4y ¼ t þ 1
þ c3 cos 3t þ c4 sin 3t
19 ð4t2 5t þ 37
9 Þ;
y ¼ c1 et c2 et þ c3 cos 3t
þ 38
þ c4 sin 3t þ 19 ð5t2 4t þ 44
9Þ
1.118
n
Engineering Mathematics-II
d2 y
dt2
dx
þ dy
dt 2y ¼ sin t; dt þ x 3y ¼ 0:
3
Ans x ¼ c1 e2t 3c2 et
2
3 t
10
e ðcos t 2sin tÞ;
(b)
:
1
y ¼ c1 et þ c2 e2t 10
ðcos t þ 3 sin tÞ
þ 3x þ 2y ¼ 2e2t :
6
Ans x ¼ c1 et þ c2 e5t þ e2t ;
7
8
5t
t
y ¼dyc2 e c1 e þ e2t
(d) dx
þy¼
sint;
þ
x¼
cost
subject
7 to x ¼ 2
dt
dt
and y ¼ 0 when t ¼ 0.
Ans. x ¼ et + et, y ¼ et – et + sin t
(c)
dx
dt
þ 2x þ 3y ¼ 0;
dy
dt
:
(e)
2
2 ddt2x
þ
3 dy
dt
¼ 4;
2
2 ddt2y
0
3 dx
dt
¼ 0 subject
0
to x(0) ¼ y(0) ¼ 0, x (0) ¼ y (0) ¼ 0.
Ans x ¼ 89 1 cos 3t2 ; y ¼ 43 t 89 sin 3t2
:
30. An uncharged condenser of capacity C is
t ffi
charged by applying an e.m.f. E sin pffiffiffiffi
LC
through leads of self inductance L and negligible resistance. Then find the charge at the
condenser plate at time
h t.
i
t ffi
t ffi
t ffi
pffiffiffiffi
pffiffiffiffi
cos pffiffiffiffi
Ans. EC
2 sin
LC
LC
LC
31. A circuit consists of an inductance L and a capacitor of capacity C in series. An alternating
e.m.f. E sin nt is applied to the circuit at time
t ¼ 0, the initial current and charge on the condenser being zero. Prove that the current at time t
2
is Lðn2nE
v2 Þ ðcos vt cos vtÞ;where CL v ¼ 1.
Et sin vt
If n ¼ v , show that current at time t is 2L .
2
Hint: Solve L ddtQ2 þ QC ¼ E sin nt under the
conditions I(0) ¼ 0, Q(0) ¼ 0. Then use
I ¼ dQ
dt :
32. A pendulum of length l hangs against a wall at
an angle h to the horizontal. Show
qffiffiffiffiffiffiffiffiffithat the time
l
of complete oscillation is 2 g sin
h:
33. A simple pendulum has a period T. When the
length of the string is increased by a small
fraction 1n of its original length, the period is T0 .
0
Show that (approximately) 1n ¼ 2ðT TT Þ :
34. How many seconds a clock would lose per day if
the length of its pendulum were increased in the
ratio 900:901.
Ans. 48 s/day
Solution about regular points
35. Solve the following differential equations about
x ¼ 20.
(i) ddx2y þ x2 y ¼ 0
h
i
x4
x8
x12
Ans y ¼ a0 x 4:3
þ 8:7:4:3
12:11:8:7:4:3
þ ...
h
i
x5
x9
x13
þ 9:8:5:4
13:12:9:8:5:4
þ ...
þ a1 x 5:4
:
2
(ii) ðx2 þ 1Þ ddx2y þ xy dy
dx ¼ xy ¼ 0
3 5
Ans y ¼ a0 1 16 x3 40
x þ ...
1 4
3 5
þ a1 x 16 x3 þ 12
x þ 40
x þ ...
:
(iii)
d2 y
dx2
2
þ x dy
dx þ ð2x þ 1Þy ¼ 0
1 4
x þ ...
Ans y ¼ a0 1 12 x2 24
3
x5
...
þ a1 x x3 30
:
2
(iv) ddx2y þ x dy
dx þ ð3x þ 2Þy ¼ 0
3
4
5
Ans y ¼ a0 1 x2 x2 þ x3 þ 11x
þ
.
.
.
40
x3
x4
x5
þ a1 x 2 4 þ 8 þ . . .
:
36. Find the power series solution in the powers of
(x – 1) of the initial value problem.
d 2 y dy
x 2 þ þ 2y ¼ 0; yð1Þ ¼ 1; y0 ð1Þ ¼ 2 :
dx dx
Hint: x ¼ 1 is the ordinary point. So consider
1
P
d2 y
an ðx 1Þn , find dy
y¼
dx and dx2 , put in the
n¼0
given equation. Use the initial conditions in
finding a0 and a1 by putting x ¼ 1 in y and dy
dx.
Other coefficients to be found by equating to
zero the powers of (x – 1).
:
Ans y ¼ 1 þ 2ðx 1Þ 2ðx 1Þ2
þ 23 ðx 1Þ3 16 ðx 1Þ4 þ . . .
Solution about regular singular point
37. Solve the following differential equations about
x¼0
2
(i) xð1 xÞ ddx2y þ ð1 xÞ dy
dx y ¼ 0
:
Ans y ¼ a0 ð1 þ log xÞð1 þ x þ 2x4
2
3
2
þ 2:5
4:9 x þ . . .Þ þ a0 ð2x x . . .Þ
n
Ordinary Differential Equations
2
(ii) xð1 xÞ ddx2y ð1 þ 3xÞ dy
dx y ¼ 0
Hint: Squaring and integrating the Jocobi series
:
cosðxsinÞ¼ J0 þ2J2 cos2þ2J4 cos4þ... ;
Ans y ¼ ðA þ B log xÞð1:2x2 þ 2:3x3
sinðxsinÞ¼ 2J1 sinþ2J3 sin3þ2J5 sin5þ... ;
þ 3:4x4 þ . . .Þ
we get
R
cos2 ðx sin Þd ¼ ðJ02 þ 2J22 þ 2J42 þ . . .Þ
þ Bð1 þ x þ 2x2 þ 11x3 þ . . .Þ
2
2
(iii) 2x2 ddx2y þ x dy
dx þ ðx 1Þy ¼ 0
x2
x4
þ 616
þ ...
Ans y ¼ c1 x 1 14
pffiffiffi
2
x4
þ c2 x 1 x2 þ 40
þ ...
0
R
:
sin2 ðx sin Þd ¼ ð2J12 þ 2J32 þ 2J52 þ . . .Þ :
0
Adding these expressions, we get the required result
43. Show that
(i) cos x ¼ J0 2J1 þ 2J4 þ . . .
1
P
¼ J0 þ 2
ð1Þn J2n ;
2
(iv) ð2x þ x3 Þ ddx2y dy
dx 6xy ¼ 0
4
Ans y ¼ c1 1 þ 3x2 þ 3x5 þ . . .
3
2
3x4
þ ...
þ c2 x2 1 þ 3x8 128
:
n¼1
(ii) sin x ¼ 2J1 2J3 þ 2J5 þ . . .
1
P
¼2
ð1Þn J2nþ1
n¼1
2
(v) 4x ddxy2 þ 2ð1 xÞ dy
dx y ¼ 0
2
3
Ans y ¼ c1 1 þ 2x! þ 22x:2 ! þ 23x:3 ! þ . . .
1
x
x2
x3
þ c2 x2 1 þ 1:3
þ 1:3:5
þ 1:3:5:7
þ ...
:
(vi)
2
x ddx2y
þ
:
dy
dx
þ 2y ¼ 0
1.119
2 2
Ans y ¼ ðc1 þ c2 log xÞ 1 2x þ ð22 x!Þ2 . . .
Hint: Put ¼ 2 in the Jocobi series
cosðxsinÞ ¼ J0 þ2J2 cos2þ2J4 cos4þ...
and
sinðx sin Þ ¼ 2J1 sin þ 2J3 sin 3
þ 2J5 sin 5 þ . . .
þ c2 ð4x 3x þ . . .Þ
2
Bessel’s Equation and Bessel Function
38. Write the general solution of the differential
equation
d y
dy 9
y ¼ 0 in terms of Jn ðxÞ:
x2 2 þ x þ x2 16
dx
dx
Ans. y ¼ c1 J34 ðxÞ þ c2 J 34 ðxÞ.
2
39. Express J5 ðxÞ in terms of J0 ðxÞand J1 ðxÞ.
12 192
72
Ans. J5 ðxÞ ¼ 384
x4 x2 1 J1 ðxÞ þ x x3 J0 ðxÞ
2
1
40. Solve x ddx2y þ 2 dy
dx þ 2 xy ¼ 0 in terms of Bessel’s
function.
i
pffiffiffih
Ans. y ¼ x c1 J12 pxffiffi2 þ þc2 J12 pxffiffi2
R
41. Show that J3 ðxÞdx ¼ J2 ðxÞ 2x J0 ðxÞ þ c
42. Show that J02 ðxÞ þ 2J12 ðxÞ þ 2J22 ðxÞ þ 2J32 ðxÞþ
... ¼ 1:
44. Show that
d 2
x 2
2
ðxÞ Jnþ1
ðxÞ :
½Jn ðxÞ ¼ ½Jn1
dx
2n
Legendre Equation and Legendre Polynomial
45. Express in the terms of Legendre polynomial
(i) 4x3 2x2 3x þ 8
Ans. 85 P3 ðxÞ 43 P2 ðxÞ 35 P1 ðxÞ þ 22
3 P0 ðxÞ
(ii) x4 þ 3x3 x2 þ 5x 2
: 358 P4ðxÞ þ 65 P3ðxÞ 212 P2 ðxÞ þ 345 P1ðxÞ
Ans
224
105 P0 ðxÞ
(iii) 4x2 – 3x + 2
Ans. 83 P2 ðxÞ 3P1 ðxÞ þ 10
3 P0 ðxÞ
46. Show that
0
0
0
Pn ðxÞ ¼ Pnþ1 ðxÞ 2xPn ðxÞ þ Pn1 ðxÞ
1.120
n
Engineering Mathematics-II
d
dx ½ð1
Hint: Use recurrence formulae
0
0
ð2n þ 1ÞPn ðxÞ ¼ Pnþ1 ðxÞ Pn1 ðxÞ
and
0
0
nPn ðxÞ ¼ xPn ðxÞ Pn1 ðxÞ
0
0
0
0
x2 Þ
d
dx Pn ðxÞ þ
n
2
nðn þ 1ÞPn ðxÞ ¼ 0 :
ðiÞ
Hint: Let y ¼ ðx 1Þ : Find
ðiiÞ
entiate once more to get ðx2 2xy1 ¼
2nðxy1 þ nÞ :
Then use Leibnitz’s theorem to get
From ðiÞ; Pn ðxÞ ¼ Pnþ1 ðxÞ Pn1 ðxÞ 2nPn ðxÞ
0
¼ Pnþ1 ðxÞ Pn1 ðxÞ 2xPn ðxÞ
dy
dx and
2
1Þ ddxy2 þ
ð1 x2 Þynþ2 2xynþ1 þ nðn þ 1Þyn ¼ 0
0
þ 2Pn1 ðxÞ
0
R1
47. Show that
1
48. Show that
R1
1
0
0
or
¼ Pnþ1 ðxÞ 2xPn ðxÞ þ Pn1 ðxÞ
Pn ðxÞdx ¼
2 if n ¼ 0
0 if n 1:
½ð1 x2 Þynþ1 þ nðn þ 1Þyn ¼ 0
or
d
dx
d
ð1 x2 Þ dx
dn
dxn
ðx2 1Þn
n
n
d
2
þ nðn þ 1Þ dx
n ðx 1Þ ¼ 0
0
2n
xPn ðxÞPn ðxÞdx ¼ 2nþ1
:
50. Expand
f ðxÞ ¼
Hint: Use integration by parts and
R1 2
2
Pn ðxÞdx ¼ 2nþ1
:
1
49. Using Rodrigue’s formula, show that Pn ðxÞ
satisfies the differential equation
differ-
0;
x;
1 < x < 0
0<x<1
in Fourier–Legendre series.
Ans. f ðxÞ ¼ 14 P0 ðxÞ þ 12 P1 ðxÞ
5
3
þ 16
P2 ðxÞ 32
P3 ðxÞ þ . . .
2
Graphs
The interest in the study of graph theory has increased
due to its applicability in so many fields like artificial
intelligence, electrical engineering, transportation
system, scheduling problems, economics, chemistry
and operations research.
2.1
DEFINITIONS AND BASIC CONCEPTS
Definition 2.1. A graph G (V, E) is a mathematical
structure consisting of two finite sets V and E. The
elements of V are called vertices (or nodes) and the
elements of E are called edges. Each edge is associated with a set consisting of either one or two vertices called its endpoints.
The correspondence from edges to endpoints is
called edge-endpoint function. This function is
generally denoted by g. Due to this function, some
authors denote graph by G (V, E, g ).
Definition 2.2. A graph consisting of one vertex and
no edges is called a trivial graph.
Definition 2.3. A graph whose vertex and edge sets
are empty is called a null graph.
Definition 2.4. An edge with just one endpoint is
called a loop or a self-loop.
Thus, a loop is an edge that joins a single endpoint to itself.
Definition 2.5. An edge that is not a self-loop is called
a proper edge.
Definition 2.6. If two or more edges of a graph G have
the same vertices, then these edges are said to be
parallel or multi-edges.
Definition 2.7. Two vertices that are connected by an
edge are called adjacent.
Definition 2.8. An endpoint of a loop is said to be
adjacent to itself.
Definition 2.9. An edge is said to be incident on each
of its endpoints.
Definition 2.10. Two edges incident on the same endpoint are called adjacent edges.
Definition 2.11. The number of edges in a graph G
which are incident on a vertex is called the degree
of that vertex.
Definition 2.12. A vertex of degree zero is called an
isolated vertex.
Thus, a vertex on which no edges are incident is
called isolated.
Definition 2.13. A graph without multiple edges (parallel edges) and loops is called simple graph.
Notations: In pictorial representations of a graph,
the vertices will be denoted by dots and edges by
line segments.
EXAMPLE 2.1
(a) Let
V {1, 2, 3, 4}
and E {e1, e2, e3, e4, e5}.
Let g be defined by
g (e1) g (e5) {1, 2}, g (e2) {4, 3},
g (e3) {1, 3}, g (e4) {2, 4}.
2.2

Engineering Mathematics-II
We note that both edges e1 and e5 have same endpoints {1, 2}. The endpoints of e2 are {4, 3}, the
endpoints of e3 are {1, 3} and endpoints of e4 are
{2, 4}. Thus the graph is as shown in the Figure 2.1.
(iv)
e5
1
e5
2
e1
1
or
e4
e3
3
3
4
2
e1
e3
e4
(ii)
(iii)
4
e2
e2
Figure 2.1
(b) For the graph pictured in the Figure 2.2,
(i) Write down the degree of the vertices A,
B, D.
(ii) Which of the edges are parallel?
(iii) Which vertex is isolated?
(iv) Point out whether it is a simple graph or a
multi-graph.
(v) Point out one pair of adjacent vertices and
one pair of adjacent edges.
(v)
The vertex D has degree 3 (because e6 is a loop
and so has degree 2)
The edges e1 and e2 are parallel
The vertex F is isolated because no edge is
incident on F
It is a multi-graph because it has multi-edges
and a loop.
A and B are adjacent vertices because they are
connected by the edge e2 or e1 whereas A and E
are not adjacent.
The edges e2 and e3 are adjacent edges because
they are incident on the same vertex B.
(c) Consider the graph with the vertices A, B, C, D
and E pictured in Figure 2.3.
A
B
C
e1
E
D
e2
A
B
Figure 2.3
e3
e4
E
C
e5
F
D
e6
Figure 2.2
In this graph,
(i) We notice that
Degree of the vertex A 2
Degree of the vertex B 4
Degree of the vertex D 3
(because loop e6 has degree 2).
In this graph, we note that
Number of edges 5
Degree of vertex A 4
Degree of vertex B 2
Degree of vertex C 3
Degree of vertex D 1
Degree of vertex E 0
Sum of the degree of vertices
4 2 3 1 0 10.
Thus, we observe that
5
∑ deg(v ) = 2e,
i
i =1
where deg(vi) denotes the degree of vertex vi and e
denotes the number of edges.

Graphs
2.3
Solution. The answer is No. In such a graph, there
will be three vertices of odd degree which is impossible.
We can also argue as follows: Total degree of
the graph (if possible) is equal to 2 2 3 6 5 7
8 4 37, which is odd. This contradicts the first
theorem of graph theory, according to which, total
degree of a graph is always even.
Euler’s Theorem 2.1. (The First Theorem
of Graph Theory)
The sum of the degrees of the vertices of a graph G is
equal to twice the number of edges in G.
Thus, total degree of a graph is even.
Proof: Each edge in a graph contributes a count of
1 to the degree of two vertices (endpoints of the
edge), that is, each edge contributes 2 to the degree
sum. Therefore, the sum of degrees of the vertices is
equal to twice the number of edges.
2.2
Corollary 2.1. There can be only an even number of
vertices of odd degree in a given graph G.
Definition 2.14. A graph G is said to simple if it has
no parallel edges or loops. In a simple graph, an
edge with endpoints v and w is denoted by {v, w}.
Proof: We know, by the fundamental theorem, that
n
∑ deg(v ) = 2 × number of edges.
i
i =1
Thus the right-hand side is an even number. Hence
to make the left-hand side an even number there can
be only even number of vertices of odd degree.
Remarks 2.1.
(i) A vertex of degree d is also called a d-valent
vertex.
(ii) The degree (or valence) of a vertex v in a graph
G is the number of proper edges incident on v
plus twice the number of self-loops.
Theorem 2.2. A nontrivial simple graph G must have
at least one pair of vertices whose degrees are equal.
Proof: Let the graph G has n vertices. Then there
appear to be n possible degree values, namely
0, 1, …, n 1. But there cannot be both a vertex
of degree 0 and a vertex of degree n 1 because
if there is a vertex of degree 0 then each of the
remaining n 1 vertices is adjacent to at most
n 2 other vertices. Hence the n vertices of G
can realize at most n 1 possible values for their
degrees. Hence, the pigeonhole principle implies
that at least two of the vertices have equal degree.
EXAMPLE 2.2
Is there a graph with eight vertices of degree 2, 2, 3,
6, 5, 7, 8, 4?
SPECIAL GRAPHS
Definition 2.15. For each integer n t 1, let Dn denote
the graph with n vertices and no edges. Then Dn is
called the discrete graph on n vertices.
For example, we have
•
•
•
and
•
•
•
•
•
D3
D5
Definition 2.16. Let n t 1 be an integer. Then a simple graph with n vertices in which there is an edge
between each pair of distinct vertices is called the
complete graph on n vertices. It is denoted by Kn.
For example, the complete graphs K2, K3 and K4
are shown in Figure 2.4.
v4
v3
v1
v2
K2
v1
K3
v2
v1
v3
K4
v2
Figure 2.4
Definition 2.17. If each vertex of a graph G has the
same degree as every other vertex, then G is called
a regular graph.
A k-regular graph is a regular graph whose
common degree is k.
For example, consider K3. The degree of each
vertex in K3 is 2. Hence K3 is regular. Similarly, K4 is
regular. Also the graph shown in Figure 2.5 is regular because degree of each vertex here is 2.
But this graph is not complete because v2 and v4
have not been connected through an edge. Similarly,
v1 and v3 are not connected by any edge.
2.4

Engineering Mathematics-II
v4
v3
,
v1
Figure 2.8
v2
2-regular graph
Figure 2.5
Thus, a complete graph is always regular but a
regular graph need not be complete.
EXAMPLE 2.5
The smallest possible simple graph that is not bipartite is the complete graph K3 shown in Figure 2.9.
EXAMPLE 2.3
The oxygen molecule O2, made up of two oxygen
atoms linked by a double bond can be represented by
the regular graph shown in Figure 2.6.
K3
Figure 2.9
Figure 2.6
Definition 2.18. Let n t 1 be an integer. Then a graph
Ln with n vertices {v1, v2, …, vn} and with edges
{vi,vi 1} for 1 d i n is called a linear graph on n
vertices.
For example, the linear graphs L2 and L4 are
shown in Figure 2.7.
It is also called path graph denoted by Pn.
v1
v2
v1
L2(or P2)
v2
v3
v4
L4(or P4)
Figure 2.7
Definition 2.19. A bipartite graph G is a graph
whose vertex set V can be partitioned into two subsets U and W, such that each edge of G has one endpoint in U and one endpoint in W.
The pair (U, W) is called a vertex bipartition of
G and U and W are called the bipartition subsets.
Obviously, a bipartite graph cannot have any selfloop.
EXAMPLE 2.4
If vertices in U are solid vertices and vertices in W
are hollow vertices, then the graphs shown in Figure
2.8 are bipartite graphs.
Definition 2.20. A complete bipartite graph G is a
simple graph whose vertex set V can be partitioned
into two subsets U {v1, v2, …, vm) and W {w1, w2,
…, wn} such that for all i, k in {1, 2, …, m} and j, l
in {1, 2, …, n}
(i) There is an edge from each vertex vi to
each vertex wj.
(ii) There is not an edge from any vertex vi to
any other vertex vk.
(iii) There is not an edge from any vertex wj to
any other vertex wl.
A complete bipartite graph on (m, n) vertices is
denoted by Km, n.
EXAMPLE 2.6
The complete bipartite graphs K3,2 and K3,4 are
shown in Figure 2.10.
w1
v1
v1
w1
v2
w2
v3
w2
v2
w3
v3
k32
k34
Figure 2.10
w4
Graphs
Definition 2.21. A cycle graph is a single vertex
with a self-loop or a simple connected graph C with
_Vc_ _Ec_, that can be drawn so that all of its vertices
and edges lie on a circle.
An n-vertex cycle graph is denoted by Cn. The cycle
graphs C1, C2, and C4 are shown in Figure 2.11.

2.5
Definition 2.24. A graph consisting of two concentric
n-cycles in which each of the n pairs of the corresponding vertices is joined by an edge is called the
circular ladder graph CLn.
For example, the graphs CL2 and CL4 are shown
in the Figure 2.14.
and
C1
C2
C4
CL2
CL4
Figure 2.11
Figure 2.14
Definition 2.22. The Petersen graph is the 3-regular
graph shown in Figure 2.12.
Definition 2.25. A graph consisting of a single vertex
having n loops is called a bouquet and is denoted
by Bn.
For example, the graphs shown in Figure 2.15
represent bouquet B2 and bouquet B4.
,
(Petersen graph)
B2
Figure 2.12
B4
Figure 2.15
This graph is frequently used to establish theorems
and to test conjectures.
Definition 2.23. The hypercube graph Qn is the n-regular graph whose vertex set is the set of bit string of
length n and such that there is an edge between two
vertices if and only if they differ in exactly one bit.
For example, 8-vertex cube graph is a hypercube
graph Q3 shown in Figure 2.13.
011
Definition 2.26. A graph consisting of two vertices
and n edges joining them is called a dipole. It is
denoted by Dn.
For example, Figure 2.16 represents D2, D3 and D4.
D2
D3
111
z-axis
D4
001
101
Figure 2.16
010
110
x-axis
y-axis 000
Q3
100
Figure 2.13
EXAMPLE 2.7
Draw all simple graphs with vertices {u, v, x, y} and
two edges, one of which is {u, v}.
Solution. In a simple graph, an edge corresponds to a
subset of two vertices. Since there are four vertices,
2.6

Engineering Mathematics-II
we can have 4c2 6 such subsets in all and they are
{u, v}, {u, x}, {u, y}, {v, x}, {v, y} and {x, y}. But one
edge of the graph is specified to be {u, v}. Hence the
possible simple graphs are as shown in Figure 2.17.
A
B
F
E
G
H
C
D
u
v
u
v
u
v
x
y
x
y
x
y
G1
G3
G2
u
v
u
y
x
x
G4
Figure 2.20
is a subgraph of the graph given in the Figure 2.21.
A
B
y
v2
e1
v1
e5
e3
e2
and
C
v2
Definition 2.28. A subgraph H is said to be a proper
subgraph of a graph G if vertex set VH of H is a
proper subset of the vertex set VG of G or edge set
EH is a proper subset of the edge set EG.
For example, the subgraphs in the above examples are proper subgraphs of the given graphs.
Definition 2.29. A subgraph H is said to span a graph
G if VH VG.
Thus H is a spanning subgraph of graph G if it
contains all the vertices of G.
For example, the subgraph (Figure 2.22)
e4
v1
v3
v4
v3
H
Figure 2.21
Definition 2.27. A graph H is said to be a subgraph
of a graph G if and only if every vertex in H is also
a vertex in G, every edge in H is also an edge in G
and every edge in H has the same endpoints as in G.
We may also say that G is a supergraph of H.
For example, the graphs (Figure 2.18).
e1
G
D
G5
SUBGRAPHS
v1
F
v
Figure 2.17
2.3
E
v2
v4
Figure 2.18
v5
are subgraphs of the graph given in Figure 2.19.
v3
v2
Figure 2.22
e1
v1
e2
v3
e5
v2
e3
e6
e4
spans the graph give in Figure 2.23.
v1
v2
v4
v5
Figure 2.19
v3
Similarly, the graph (Figure 2.20)
v4
Figure 2.23
Graphs
Definition 2.30. Let G (V, E) be a graph. Then
the complement of a subgraph Gc (V c, Ec) with
respect to the graph G is another subgraph Gs (Vs,
Es) such that Es E Ec and V s contains only the
vertices with which the edges in Es are incident.
For example, the subgraph (Figure 2.24)

2.7
v2
v1
v3
v4
G
Figure 2.27
v1
v2
Figure 2.24
Then complement Gc of G is the graph shown in
Figure 2.28.
v2
is the complement of the subgraph (Figure 2.25).
v1
v1
v7
v8
v3
v2
v4
v3
v5
v6
G′
Figure 2.28
v4
Figure 2.25
with respect to the graph G shown in Figure 2.26.
EXAMPLE 2.8
Find the complement of the graphs:
(a)
v2
v1
v7
v8
v2
v1
v3
v5
v6
v3
v4
v4
Figure 2.29
Figure 2.26
Definition 2.31. If G is a simple graph, the complement of G (edge complement), denoted by Gc or Gc
is a graph such that
(i) The vertex set of Gc is identical to the vertex set of G, that is, VGc VG.
(ii) Two distinct vertices v and w of Gc are connected by an edge if and only if v and w
are not connected by an edge in G.
For example, consider the graph G shown in
Figure 2.27.
(b)
(c)
v1
v2
v3
v4
Complete graph K4:
v1
v2
v3
v4
Figure 2.31
2.8

Engineering Mathematics-II
Solution.
(a)
v1
v2
v1
v3
v2
v3
v4
v5
v4
v6
Figure 2.32
GC
Figure 2.35
(b)
v1
v2
v3
v4
Figure 2.33
(c)
Null graph.
Definition 2.32. If a new vertex v is joined to each
of the pre-existing vertices of a graph G, then the
resulting graph is called the join of G and v or the
suspension of G from v. It is denoted by G v.
Thus, a graph obtained by joining a new vertex v
to each of the vertices of a given graph G is called
the join of G and v or the suspension of G from v.
For example, consider the graph given below
(Figure 2.36).
EXAMPLE 2.9
Find the edge complement of the graph G shown in
Figure 2.34.
v1
v2
v3
v4
Figure 2.36
v1
v2
v3
v4
v5
v6
Let v be a vertex. Then the graph shown in the Figure 2.37 is the join of G to v.
v
v1
v2
G
Figure 2.34
Solution. The edge complement of G is the following
graph Gc (Figure 2.35).
v3
v4
G+v
Figure 2.37
Graphs
2.4
ISOMORPHISMS OF GRAPHS
We know that shape or length of an edge and its
position in space are not part of specification of a
e1
v1
2.9
graph. For example, the Figures 2.38 represent the
same graph.
v3
v2

e2
v1
e1
v2
and
e3
e2
e3
v3
Figure 2.38
Definition 2.33. Let G and H be graphs with vertex sets V(G) and V(H) and edge sets E(G) and
E(H), respectively. Then G is said to isomorphic
to H if there exist one-to-one correspondences
g: V(G) o V(H) and h: E(G) o E(H) such that for all
v  V(G) and e  E(G),
v is an endpoint of e œ g(v) is an endpoint of h(e).
Definition 2.34. The property of mapping endpoints
to endpoints is called preserving incidence or the
continuity rule for graph mappings.
As a consequence of this property, a self-loop
must map to a self-loop.
Thus, two isomorphic graphs are same except for
the labelling of their vertices and edges. For example,
the graphs shown in the Figure 2.39 are isomorphic.
a′
a
b
and
c
c′
d
d′
b′
Figure 2.39
Similarly, the hypercube graph Q3 and circular ladder CL4 shown in Figure 2.40 are isomorphic.
f (010)
011
001
111
101
010
000
f (011)
f (000)
110
100
f (001)
f (111)
f (101)
f (100)
Hypercube Q3
Circular ladder CL4
Figure 2.40
f (110)
2.10

Engineering Mathematics-II
V(G)
EXAMPLE 2.10
Show that the graphs shown in Figure 2.41 are
isomorphic.
v2
e7
e5 e6
v5
v1
e2
e3
and
h
e4
v4
e1
e2
e3
e4
e5
w1
f3
f1
w1
w2
w3
w4
w5
v1
v2
v3
v4
v5
v3
e1
V(H)
g
w2 f4
f2
f6
f7
w3
w4
f1
f2
f3
f4
f5
f6
e6
e7
w5
H
f7
Figure 2.41
Figure 2.42
Solution. To solve this problem, we have to find g:
V(G) o V(H) and h: E(G) o E(H) such that for all v
 V(G) and e  E(G),
v is an endpoint of e œ g(v) is an endpoint of h(e).
Since e2 and e3 are parallel (have the same endpoints), h(e2) and h(e3) must also be parallel. Thus
we have
h(e2) f1 and h(e3) f2
or h(e2) f2 and h(e3) f1.
Also the endpoints of e2 and e3 must correspond to
the endpoints of f1 and f2 and so
g(v3) w1 and g(v4) w5 or
g(v4) w1.
g(v3) w5 and
Further, we note that v1 is the endpoint of four distinct edges e1, e7, e5 and e4 and so g(v1) should be
the endpoint of four distinct edges. We observe that
w2 is the vertex having four edges and so g(v1) w2.
If g(v3) w1, then since v1 and v3 are endpoints of e1
in G, g(v1) w2 and g(v3) w1 must be endpoints of
h(e1) in H. This implies that h(e1) f3.
Continuing this way, we can find g and h to define
the isomorphism between G and H.
One such pair of functions (of course there exist
several) is shown in Figure 2.42.
Remark 2.2. Each of the following properties of a
graph is invariant under graph isomorphism, where
n, m and k are all non-negative integers:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Has n vertices
Has m edges
Has a vertex of degree k
Has m vertices of degree k
Has a circuit of length k
Has a simple circuit of length k
Has m simple circuits of length k,
Is connected
Has an Euler circuit
Has a Hamiltonian circuit
EXAMPLE 2.11
Examine for isomorphism
(a)
and
G
H
⎫
⎪
⎪⎪
⎬
⎪
⎪
⎪⎭
to be
studied
later on
in this
chapter

Graphs
2.11
Thus a simple path is a walk of the form
(b)
{v0, e1,v1, e2, v2, …, vi 1, en, vn},
and
where all the ei and all the vi are distinct.
G
H
(c)
Definition 2.38. A walk in a graph G that starts and
ends at the same vertex is called a closed walk.
Definition 2.39. A closed walk that does not contain a
repeated edge is called a circuit.
Thus, a closed path is called a circuit (or a cycle)
if it is a walk of the form
{v0, e1, v1, e2, v2, …,vn 1, en, vn},
and
where v0 vn and all the ei are distinct.
G
H
Figure 2.43
Solution. We note in Figure 2.43 that
(a) G has nine edges whereas H has only eight.
Hence, G is not isomorphic to H.
(b) G has a vertex v of degree 4, whereas H has no
vertex of degree 4. Hence, G is not isomorphic
to H.
(c) G has a simple circuit of length 3 (marked
dark) whereas H does not have. Hence, G is
not isomorphic to H.
2.5
WALKS, PATHS AND CIRCUITS
Definition 2.35. In a graph G, a walk from vertex v0 to
vertex vn is a finite alternating sequence {v0, e1, v1, e2,
…, vn 1, en, vn} of vertices and edges such that vi 1
and vi are the endpoints of ei.
The trivial walk from a vertex v to v consists of
the single vertex v.
Definition 2.36. In a graph G, a path from the vertex
v0 to the vertex vn is a walk from v0 to vn that does not
contain a repeated edge.
Thus a path from v0 to vn is a walk of the form
{v0, e1, v1, e2, v2, …, vn 1, en, vn},
where all the edges ei are distinct.
Definition 2.37. In a graph, a simple path from v0 to
vn is a path that does not contain a repeated vertex.
Definition 2.40. A simple circuit is a circuit that does
not have any other repeated vertex except the first
and the last.
Thus, a simple circuit is a walk of the form
{v0, e1, v1, e2, …, vn 1, en, vn},
where all the ei are distinct and all the vj are distinct
except that v0 vn.
EXAMPLE 2.12
In the graph given in the Figure 2.44, determine
whether the following walks are paths, simple paths,
closed walks, circuits, simple circuits or are just
walks.
1.
2.
3.
4.
5.
6.
v1 e2 v2 e3 v3 e4 v4 e5 v2 e2 v1 e1 v0,
v5 v4 v2 v1,
v2 v3 v4 v5 v2,
v4 v2 v3 v4 v5 v2 v4,
v2 v1 v5 v2 v3 v4 v2,
v0 v5 v2 v3 v4 v2 v1.
v0
e1
e10
v1
e9
v2
e7
e8
v5
v3
e3
e2
e6
e5
e4
v4
Figure 2.44
Solution
1. It is just a walk because it has repeated edges
and repeated vertices.
2.12

Engineering Mathematics-II
2. It is a simple path, because here no vertex is
repeated and no edge is repeated.
3. It is closed walk in which no edge is repeated
and hence it is a circuit. It starts and ends at the
same vertex, and therefore is a simple circuit.
4. It is a closed walk in which no edge is repeated
but vertices v2 and v4 are repeated. Hence it is a
circuit.
5. It is a closed walk in which no edge is repeated.
Hence it is a circuit. Only one vertex is repeated
twice, hence it is not a simple circuit.
6. It is a walk in which no edge is repeated but the
vertex v2 is repeated. Hence it is a path.
EXAMPLE 2.13
Consider the graph shown in Figure 2.45.
e1
v2
v1
e2
Figure 2.46
EXAMPLE 2.14
Consider the graph shown in Figure 2.47.
v1
e1
e5
e6
v5
v2
e2
v3
e7
e3
e4
v4
e8
Figure 2.47
e1
e2
v1
e4
v2
e3
In this graph, we observe that
(i) The path v4 e8 v4 is a cycle of length 1.
(ii) The path v5 e4 v4 e8 v4 e3 v3 from v5 to v3 (Figure
2.48) is a path of length 3.
v3
v3
e3
e5
v5
e4
v4
e8
v4
Figure 2.45
Figure 2.48
We note that e3, e5 is a path. The walk e1, e2, e3, e5 is
a path but it is not a simple path because the vertex
v1 is repeated (e1 being a self-loop). The walk e2, e3,
e4 is a circuit. The walk e2, e3, e4, e1 is a circuit but it
is not simple circuit because vertex v1 repeats twice
(or we may write that v1 is met twice).
(iii) The path v1 e1 v2 e6 v5 e5 v1 (Figure 2.49) is a
Definition 2.41. In a graph the number of edges in
the path {v0, e1, v1, e2,…, en, vn} from v0 to vn is
called the length of the path.
Definition 2.42. A cycle with k-edges is called a kcycle or cycle of length k.
For example, loop is a cycle of length 1. On the
other hand, a pair of parallel edges e1 and e2, shown
in the Figure 2.46, is a cycle of length 2.
v1
e5
e1
v2
e6
v5
Figure 2.49
cycle of length 3, since it consists of three edges.
Definition 2.43. A graph is said to be acyclic if it contains no cycle.
For example, the graphs shown in Figure 2.50 are
acyclic.

Graphs
(b)
2.13
v3
v4
v1
and
v2
v5
v6
Figure 2.50
(c)
Theorem 2.3. If there is a path from vertex v1 to v2 in a
graph with n vertices, then there does not exist a simple path of more than n 1 edges from vertex v1 to v2.
Proof: Suppose there is a path from v1 to v2. Let
v1, …, vi, …, v2 be the sequence of vertices which
the path meets between the vertices v1 and v2. Let
there be m edges in the simple path. Then there
will be m 1 vertices in the sequence. Therefore
if m ! n 1, then there will be more than n vertices
in the sequence. But the graph is with n vertices.
Therefore some vertex, say vk, appears more than
once in the sequence. So the sequence of vertices
shall be v1, …, vi, …, vk, …, vk, …, v2. Deleting the
edges in the path that lead vk back to vk we have
a path from v1 to v2 that has fewer edges than the
original one. This argument is repeated until we
get a path that has n 1 or fewer edges.
Definition 2.44. Two vertices v1 and v2 of a graph G
are said to be connected if and only if there is a walk
from v1 to v2.
Definition 2.45. A graph G is said to be connected
if and only if given any two vertices v1 and v2 in G,
there is a walk from v1 to v2.
Thus, a graph G is connected if there exists a
walk between every two vertices in the graph.
v1
v2
v3
v4
v1
v3
v6
v2
v4
Figure 2.51
Solution. In Figure 2.51, we observe that
(a) Since every two vertices in the given graph are
connected by a path, therefore the given graph
is connected.
(b) This graph is also connected.
(c) The graph in (c) is disconnected. It has two connected components.
(d) In this graph, the edge (v5, v6) cross the edges
(v1, v2) and (v3, v4) at points which are not vertices. Therefore the graph can be redrawn as
shown in Figure 2.52.
v1
v2
v3
v4
v6
v3
v5
v4
v5
v7
v5
EXAMPLE 2.15
Which of the graphs shown below are connected?
(a)
v2
v6
(d)
Definition 2.46. A graph which is not connected is
called disconnected graph.
v1
v5
Figure 2.52
2.14

Engineering Mathematics-II
There is no path from v2 to v4, etc. Hence the given
graph is disconnected and has three connected components.
Solution
(a) The graph in (a) has four connected components (Figures 2.55 and 2.56)
EXAMPLE 2.16
Which of the graphs shown in Figure 2.53 are connected?
(a)
v1
e1
v3
e2
v2
(b)
v1
v2
Figure 2.55
v1
v2
v3
v3
H1: with vertex set {v1, v2, v3} and edge set {e1, e2}
H2: with vertex set {v4} and edge set f
H3: with vertex set {v5} and edge set f
H4: with vertex set {v6, v7, v8,v9} and edge set
{e3, e4, e5}
v4
v6
v4
v8
e3
e5
v5
Figure 2.53
e4
v9
Solution. Graph (a) is not connected as there is no
walk from any of v1, v2, v3, v4 to the vertex v5.
The graph (b) is clearly connected.
Definition 2.47. If a graph G is disconnected, then the
various connected pieces of G are called the connected components of the graph.
v7
Figure 2.56
(b) The graph in (b) is disconnected and have two
connected components (Figures 2.57(a), 2.57(b))
H1:
e1
v1
v2
EXAMPLE 2.17
Find the connected components of the graphs given
below (Figure 2.54).
e2
(a)
Figure 2.57(a)
v1
e1
v3
e2
v4
v6
e3
v9
e4
v5
v3
v8
e5
v2
e3
v7
with vertex set {v1, v2, v3} and edge set {e1, e2, e3}
H2:
v5
e4
(b)
v5
e1
v1
e2
e4
v2
e3
v4
e5
e6
v3
Figure 2.54
v4
v6
e5
e6
v6
Figure 2.57(b)
Graphs

2.15
v1
with vertex set {v4, v5, v6} and edge set {e4,
e5, e6}.
e1
v2
EXAMPLE 2.18
Find the number of connected components in the
graph shown below (Figure 2.58).
e2
e5
v5
v3
e3
e4
v4
Figure 2.60
(a) v1 e1 v2 e2 v3 e3 v4 e4 v5 e5 v2
(b) v1 e1 v2 e2 v3 e3 v4 e4 v5 e5 v2 e1 v1
(c) v2 e2 v3 e3 v4 e4 v5 e5 v2
Figure 2.58
Solution. The connected components are shown in
Figure 2.59.
and
Figure 2.59
Remark 2.3. If a connected component has n vertices, then degree of any vertex cannot exceed n 1.
2.6
EULERIAN PATHS AND CIRCUITS
Definition 2.48. A path in a graph G is called an Euler
path if it includes every edge exactly once.
Definition 2.49. A circuit in a graph G is called an
Euler circuit if it includes every edge exactly once.
Thus, an Euler circuit (Eulerian trail) for a graph G
is a sequence of adjacent vertices and edges in G
that starts and ends at the same vertex, uses every
vertex of G at least once, and uses every edge of
G exactly once.
Definition 2.50. A graph is called Eulerian graph if
there exists an Euler circuit for that graph.
EXAMPLE 2.19
Find which of the following are Euler paths and Euler
circuits in the graph given below (Figure 2.60)
Solution
(a) The walk in (a) is an Euler path but it is not
Euler circuit because it is not closed.
(b) It is not an Euler circuit because the edge e1 is
covered twice.
(c) It is neither an Euler path nor an Euler circuit
because the vertex v1 of G has not been used.
Theorem 2.4. If a graph has an Euler circuit, then
every vertex of the graph has even degree.
Proof: Let G be a graph which has an Euler circuit.
Let v be a vertex of G. We shall show that degree of
v is even. By definition, Euler circuit contains every
edge of graph G. Therefore the Euler circuit contains
all edges incident on v. We start a journey beginning
in the middle of one of the edges adjacent to the start
of Euler circuit and continue around the Euler circuit to end in the middle of the starting edge. Since
Euler circuit uses every edge exactly once, the edges
incident on v occur in entry/exist pair and hence the
degree of v is a multiple of 2. Therefore, the degree
of v is even. This completes the proof of the theorem.
Starting point
v
Figure 2.61
We know that contrapositive of a conditional
statement is logically equivalent to the statement.
Thus, Theorem 2.4 is equivalent to the following:
Theorem 2.5. If a vertex of a graph is not of
even degree, then it does not have an Euler circuit.
2.16

Engineering Mathematics-II
Thus,
If some vertex of a graph has odd degree, then
that graph does not have an Euler circuit.
Theorem 2.6. If G is a connected graph and every
vertex of G has even degree, then G has an Euler
circuit.
EXAMPLE 2.20
Show that the graphs shown in Figure 2.62 do not
have Euler circuits.
Proof: Let every vertex of a connected graph G has
even degree. If G consists of a single vertex v, the
trivial walk from v to v is an Euler circuit. So suppose that G consists of more than one vertices. We
start from any vertex v of G. Since the degree of
each vertex of G is even, if we reach each vertex
other than v by travelling on one edge, the same vertex can be reached by travelling on another previously unused edge. Thus a sequence of distinct adjacent edges can be produced indefinitely as long as
v is not reached. Since the number of edges of the
graph is finite (by definition of graph), the sequence
of distinct edges will terminate. Thus the sequence
must return to the starting vertex. We thus obtain a
sequence of adjacent vertices and edges starting and
ending at v without repeating any edge. Thus we get
a circuit C.
If C contains every edge and vertex of G, then C
is an Euler circuit.
If C does not contain every edge and vertex of G,
remove all edges of C from G and also any vertices
that become isolated when the edges of C are
removed. Let the resulting subgraph be Gc. We note
that when we removed edges of C, an even number
of edges from each vertex have been removed. Thus
degree of each remaining vertex remains even.
Further, since G is connected, there must be at
least one vertex common to both C and Gc. Let it
be w (in fact there are two such vertices). Pick any
sequence of adjacent vertices and edges of Gc starting and ending at w without repeating an edge. Let
the resulting circuit be Cc.
Join C and Cc together to create a new circuit Cs.
Now, we observe that if we start from v and follow C
all the way to reach w and then follow Cc all the way
to reach back to w. Then continuing travelling along
the untravelled edges of C, we reach v.
If Cs contains every edge and vertex of C, then
Cs is an Euler circuit. If not, then we again repeat
our process. Since the graph is finite, the process
must terminate.
The process followed has been described in the
graph G shown below (Figure 2.64).
(a)
(b)
v1
v2
v1
v2
v3
v4
v4
v3
Figure 2.62
Solution. In graph (a), degree of each vertex is 3.
Hence this does not have an Euler circuit.
In graph (b), we have
deg(v2) 3
and
deg(v4) 3.
Since there are vertices of odd degree in the given
graph, therefore it does not have an Euler circuit.
Remark 2.4. The converse of Theorem 2.4 is not true.
There exist graphs in which every vertex has even
degree but the Euler circuits do not exist.
For example,
and
Figure 2.63
are graphs (Figure 2.63) in which each vertex has
degree 2 but these graphs do not have Euler circuits
since there is no path which uses each vertex at
least once.
Graphs
C′
w
u
u
e
v
C
2.17
Proof: Let u and v be two vertices of odd degree in
the given connected graph G (see Figure 2.66)
G′
v

v
(Graph G)
v
w
G
C′′
G′
Figure 2.66
Figure 2.64
Theorems 2.4 and 2.6 taken together imply:
Theorem 2.7 (Euler’s Theorem). A finite connected
graph G has an Euler circuit if and only if every vertex of of G has even degree.
Thus, finite connected graph is Eulerian if and only
if each vertex has even degree.
Theorem 2.8. If a graph G has more than two vertices
of odd degree, then there can be no Euler path in G.
Proof: Let v1, v2 and v3 be vertices of odd degree.
Since each of these vertices had odd degree, any
possible Euler path must leave (arrive at) each of v1,
v2, v3 with no way to return (or leave). One vertex of
these three vertices may be the beginning of Euler
path and another the end, but this leaves the third vertex at one end of an untravelled edge. Thus, there is
no Euler path.
v1
v2
v1
v2
v3
or
v3
(Graphs having more than two vertices of odd degree).
Figure 2.65
Theorem 2.9. If G is a connected graph and has
exactly two vertices of odd degree, then there is an
Euler path in G. Further, any Euler path in G must
begin at one vertex of odd degree and end at the
other.
If we add the edge e to G, we get a connected graph
Gc all of whose vertices have even degree. Hence
there will be an Euler circuit in Gc. If we omit e
from Euler circuit, we get an Euler path beginning
at u (or v) and ending at v (or u).
EXAMPLE 2.21
Has the graph given in Figure 2.67 an Eulerian path?
B
A
C
D
Figure 2.67
Solution. In the given graph,
deg(A) 1, deg(B) 2,
deg(C) 2, deg(D) 3.
Thus the given connected graph has exactly two vertices of odd degree. Hence, it has an Eulerian path.
If it starts from A (vertex of odd degree), then it
ends at D (vertex of odd degree). If it starts from D
(vertex of odd degree), then it ends at A (vertex of
odd degree).
But, on the other hand, let the graph be as shown
below (Figure 2.68)
A
e1
B
e2
e4
D
e3
Figure 2.68
C
2.18

Engineering Mathematics-II
Here, deg(v1) 4, deg(v2) 4, deg(v3) 2, deg(v4) 2.
Thus degree of each vertex is even. But the graph is
not Eulerian since it is not connected.
Then deg(A) 1, deg(B) 3, deg(C) 1, degree of
D 3 and so we have four vertices of odd degree.
Hence it does not have an Euler path.
EXAMPLE 2.22
Does the graph given in Figure 2.69 possess an
Euler circuit?
e3
v4
EXAMPLE 2.24
Is it possible to trace the graph in Figure 2.71 without lifting the pencil?
e7
v3
v1
e4
e5
v1
e2
e6
v2
e1
v5
v8
Figure 2.69
v3
v7
Figure 2.71
Solution. The given graph is connected. Further
deg(v1) 3,
deg(v3) 3,
v9
v6
v4
v2
deg(v2) 4,
deg(v4) 4.
Solution. The problem is equivalent to say that “Is it
possible for this graph to have an Eulerian circuit”?
We observe that
Since this connected graph has vertices with odd
degree, it cannot have Euler circuit. But this graph
has Euler path, since it has exactly two vertices of
odd degree. For example, v3 e2 v2 e7 v4 e6 v2 e1 v1 e4 v4
e3 v3 e5 v1 is an Euler path.
deg(v1) 3, deg(v2) 2, deg(v3) 2, deg(v4) 3,
deg(v5) 4,
deg(v6) 2, deg(v7) 2, deg(v8) 4, deg(v9) 2.
EXAMPLE 2.23
Consider the graph given below (Figure 2.70):
v2
v3
Since the graph contains vertex of odd degree,
therefore it cannot have Euler’s circuit and therefore
can not be traced without lifting the pencil.
v1
v4
EXAMPLE 2.25
Find the Euler’s circuit for the graph given below
(Figure 2.72):
Figure 2.70
v2
e1
v1
e13
e14
v12
v11
e16
v4
e2
e12
v10
e15
e11
e3
v6
e4
v3
e5
e6
v5
e10
e9
v9
Figure 2.72
v7
e7
e8
v8
Graphs
Solution. The given connected graph has 12 vertices
and degree of each vertex is even. Hence, by Euler’s
theorem, this has an Euler’s circuit. For example, we
observe that v1 e1 v2 e13 v11 e12 v10 e15 v12 e14 v2 e2 v3
e3 v4 e4 v5 e5 v6 e6 v7 e7 v8 e8 e9 v9 e10 v3 e11 v10 e16 v1 is
an Euler’s circuit. In short we can represent it by e1,
e13, e12, e15, e14, e2, e3, e4, e5, e6, e7, e8, e9, e10, e11, e16.
EXAMPLE 2.26 (The bridges of Konigsberg)
The graph theory began in 1736 when Leonhard
Euler solved the problem of seven bridges on Pregel River in the town of Konigsberg in Prussia (now
Kaliningrad in Russia). The two islands and seven
bridges are shown in Figure 2.73.
Island
A
Bridge
Bridge
C
D
Bridge
Bridge
Bridge
Island
B
Bridge

2.19
Thus the graph of the problem is as shown in
Figure 2.74.
A (Island)
C (Side of the river)
(Side of the river) D
B (Island)
(Euler’s graphical representation of seven bridges problem)
Figure 2.74
The problem then reduces to
“Is there any Euler’s path in the above diagram?”.
To find the answer, we note that there are more
than two vertices having odd degree. Hence there
exists no Euler path for this graph.
EXAMPLE 2.27
The floor plan shown in Figure 2.75 is for a house
that is open for public viewing. Each room is connected to every room with which it has a common
wall and to the outside along each wall. Is it possible
to begin in a room or outside and take a walk which
goes through each door exactly once?
Bridge
e1
River
D
e7
The people of Konigsberg posed the following
question to famous Swiss Mathematician Leonhard
Euler:
“Beginning anywhere and ending any where,
can a person walk through the town of Konigsberg
crossing all the seven bridges exactly once?”
Euler showed that such a walk is impossible. He
replaced the islands A, B and the two sides (banks)
C and D of the river by vertices and the bridges as
edges of a graph. We note then that
deg(A) 3, deg(B) 5,
deg(C) 3, deg(D) 3.
e3
e10
Room A
e2
Figure 2.73
D
Room B
e6
e5
e4
D
e9
Doorway
Room C
e8
D (Outside)
Figure 2.75
If each room and out side constitute a vertex and
each door corresponds to an edge, then the floor
plan converts into the graph. There are two edges
e1 and e2 from A to D, two edges e3 and e4 from B
to D, three edges e7, e8, e9 from C to D, one edge
e10 from A to B, one edge e5 from A to C, one edge
e6 from B to C. Hence the graph is as shown below
(Figure 2.76):
2.20

Engineering Mathematics-II
e1
e10
e1
A
e3
e5
v1
B
v4
v2
e2
e4
e2
e5
v3
D
e9 e e8
7
e4
v5
Figure 2.78
e6
Hence the edge e3 is a bridge in the given graph.
C
Figure 2.76
2.6.1
We note that
deg(A) deg(B) 4,
deg(C) 5,
deg(D) 7.
Since the graph is connected and degree of exactly
two vertices C and D is odd, there exists an Euler’s
path and that path should start from one vertex
of odd degree and end to the other vertex of odd
degree. Therefore, either the path will begin from C
and end at D or it will begin from D and end at C. An
Euler’s path is shown (which begins from C).
Definition 2.51. An edge in a connected graph is
called a bridge or a cut edge if deleting that edge
creates a disconnected graph.
For example, consider the graph shown below
(Figure 2.77):
Methods for Finding Euler Circuit
Method 1. We know that if every vertex of a nonempty connected graph has even degree, then the
graph has an Euler circuit. We shall make use of
this result to find an Euler path in a given graph.
Consider the graph shown in Figure 2.79.
v2
e1
v1
v6
e2
e3
e4
e9
v3
e6
e5
e8
v5
e7
v7
e11
e10
v8
v4
e12
Figure 2.79
We note that
e1
e5
v1
e3
v2
e2
v4
v3
e4
deg(v2) deg(v4) deg(v6) deg(v8) 2,
deg(v1) deg(v3) deg(v5) deg(v7) 4.
Hence all vertices have even degree. Also the given
graph is connected. Hence the given graph has an
Euler circuit. We start from the vertex v1 and let C be
C: v1 v2 v3 v1.
v5
Figure 2.77
In this graph, if we remove the edge e3, then the
graph breaks into two connected components given
below (Figure 2.78):
Then C is not an Euler circuit for the given graph but
C intersects the rest of the graph at v1 and v3.
Let Cc be
Cc: v1v4 v3 v5 v7 v6 v5 v8 v7 v1.
(In case we start from v3, then Cc will be v3 v4 v1 v7
v6 v5 v7 v8 v5).
Graphs
Path Cc into C and obtain
2.21
Let G be the connected graph as shown in the
Figure 2.80.
Cs: v1v2 v3 v1 v4 v3 v5 v7 v6 v5 v8 v7 v1,
that is,
v1
Cs: e1e2 e3 e4 e5 e6 e7 e8 e9 e10 e11 e12.
e1
Method 2: Fleury’s Algorithm: This algorithm is
used to find Euler’s circuit for a connected graph with
no vertices of odd degree.
We shall illustrate the method with the help of the
following example:
Current Path
Next Edge
e9
e8
v2
(If we had started from v2, then Cs: v1v2 v3 v4 v1 v7 v6
v5 v7 v8 v5 v3 v1 or e1e2 e5 e4 e12 e8 e9 e7 e11 e10 e6 e3).
In Cs all edges are covered exactly once. Also every
vertex has been covered at least once. Hence Cs is
an Euler circuit.
e7
e2
v3
e6
e4
e3
v6
v4
Figure 2.80
Step 1. We begin from any vertex, say v1
Step 2. We then construct the following table:
Reason
{v1, v2}
No edge from v1 is a bridge, so
choose any edge, say {v1, v2}
v 1 v2
{v2, v4}
(v2, v5) a bridge, so choose {v2, v4}
v1 v2 v4
{v4, v5}
only one edge {v4, v5}from v4
remains
v1 v2 v4 v5
{v5, v6}
Since {v2, v5} and {v5, v3} are
bridges, choose {v5, v6}
v1 v2 v4 v5 v6
{v6, v3}
only one edge {v6, v3} remains
v1 v2 v4 v5 v6 v3
{v3, v5}
Since {v1, v3} is a bridge, choose
either {v3, v2}or {v3, v5}
v1 v2 v4 v5 v6 v3 v5
{v5, v2}
only one edge {v5, v2} remains
v1 v2 v4 v5 v6 v3 v5v2
{v2, v3}
only one edge {v2, v3} remains
v1 v2 v4 v5 v6 v3 v5 v2 v3
{v3, v1}
only one edge {v3, v1} remains
C: v1 v2 v4 v5 v6 v3 v5 v2 v3 v1.
In term of edges, this Euler circuit can be expressed as
e1 e2 e3 e4 e5 e6 e7 e8 e9.
HAMILTONIAN CIRCUITS
Definition 2.52. A Hamiltonian path for a graph
G is a sequence of adjacent vertices and distinct
e5
v5
v1
Hence one possible Euler circuit is
2.7

edges in which every vertex of G appears exactly
once.
Definition 2.53. A Hamiltonian circuit for a graph G
is a sequence of adjacent vertices and distinct edges
in which every vertex of G appears exactly once,
except for the first and the last which are the same.
Definition 2.54. A graph is called Hamiltonian if it
admits a Hamiltonian circuit.
2.22

Engineering Mathematics-II
EXAMPLE 2.28
The wooden graph shown in Figure 2.81 and constructed by William Hamilton in the shape of a regular dodecahedron is a Hamiltonian circuit.
v1
Figure 2.84
v20
v19
v13
v2
v11
v3
v5
v14 v18
v12
v15
v10
v16
v6
v17
v9
v7
v8
v4
Figure 2.81
The Hamilton circuit is v1 v2 v3 … v18 v19 v20 v1.
EXAMPLE 2.29
A complete graph Kn has a Hamiltonian circuit.
In particular, the graphs shown in Figure 2.82 are
Hamiltonian.
and
K3
K4
Figure 2.82
EXAMPLE 2.30
The graph shown below does not have a Hamiltonian circuit.
v4
v1
v3
v2
v5
Figure 2.83
EXAMPLE 2.31
The graph shown in Figure 2.84 does not have a
Hamiltonian circuit
Remark 2.5. It is clear that only connected graphs
can have Hamiltonian circuit. However, there is
no simple criterion to tell us whether or not a given
graph has Hamiltonian circuit. The following results
give us some sufficient conditions for the existence
of Hamiltonian circuit/path.
Theorem 2.10. Let G be a linear graph of n vertices.
If the sum of the degrees for each pair of vertices in
G is greater than or equal to n 1, then there exists a
Hamiltonian path in G.
Theorem 2.11. Let G be a connected graph with n vertices. If n t 3 and deg(v) t n for each vertex v in G,
then G has a Hamiltonian circuit.
Theorem 2.12. Let G be a connected graph with n
vertices and let u and v be two vertices of G that
are not adjacent. If deg(u) deg(v) t n, then G has a
Hamiltonian circuit.
Corollary 2.2. Let G be a connected graph with
n vertices. If each vertex has degree greater
than or equal to n/2, then G has a Hamiltonian
circuit.
Proof: It is given that degree of each vertex is
greater than or equal to n/2. Hence the sum of the
degree of any two vertices is greater than or equal to
n/2 n/2 n. So, by the above theorem, the graph G
has a Hamiltonian circuit.
Theorem 2.13. Let n be the number of vertices and m
be the number of edges in a connected graph G. If
1
m t (n2 – 3n 6), then G has a Hamiltonian circuit.
2
The following example shows that the above conditions are not necessary for the existence of Hamiltonian path.
Graphs
EXAMPLE 2.32
Let G be the connected graph shown in the
Figure 2.85.
v7
2.23
EXAMPLE 2.33
Consider the graph G shown below (Figure 2.86):
v3
v1
v8

v6
v2
v1
v5
v5
v4
Figure 2.86
v2
v3
v4
Figure 2.85
We note that
n Number of vertices in G 8,
m Number of edges in G 8,
Degree of each vertex 2.
Thus, if u and v are non-adjacent vertices, then
deg(u) deg(v) 2 2 4 ≥/ 8.
Also,
1 2
1
(n – 3n 6)
(64 – 24 6) 23.
2
2
Clearly m ≥/ 1 (n2 – 3n 6). Therefore the
2
above two theorems fail. But the given graph has
Hamiltonian circuit. For example, v1, v2, v3, v4, v5, v6,
v7, v8, v1 is an Hamiltonian circuit for the graph.
Proposition 2.1. Let G be a graph with at least two
vertices. If G has a Hamiltonian circuit, then G has a
subgraph H with the following properties:
1.
2.
3.
4.
H contains every vertex of G
H is connected
H has the same number of edges as vertices
Every vertex of H has degree 2
The contrapositive of this proposition is
“If a graph G with at least two vertices does not have
a subgraph H satisfying (1) – (4), then G does not
have a Hamiltonian circuit”.
Also we know that contrapositive of a statement
is logically equivalent to the statement. Therefore
the above result can be used to show non-existence
of a Hamiltonian circuit.
Let us verify whether this graph has a Hamiltonian
circuit.
Suppose that G has a Hamiltonian circuit. Then it
has a subgraph H such that
1. H contains every vertex of G, i.e., H has 5 vertices v1, v2, v3, v4, v5
2. H is connected
3. H has 5 edges
4. Every vertex of H has degree 2
Since the degree of v2 in G is 4 and every vertex
of H has degree 2, two edges incident on v2 must
be removed from G to create H. But we note that
the edge {v1, v2} cannot be removed, because the
removal of {v1, v2} will decrease the degree of v1 to
1, which is less than 2. Similarly the edge {v2, v3},
{v2, v4} and {v2, v5} cannot be removed. Hence the
degree of v2 in H must stand at 4 which contradicts
the condition that every vertex in H has degree 2 in
H. Therefore, no such subgraph H exists. Hence G
does not have a Hamiltonian circuit.
EXAMPLE 2.34
Does the graph G given below (Figure 2.87) have
Hamiltonian circuit?
a
b
e
c
d
Figure 2.87
Solution. The given graph has
n Number of vertices 5
m Number of edges 8
2.24

Engineering Mathematics-II
deg(a) deg(b) deg(c) deg(d) 3
deg(e) 4
We observe that
(i) Degree of each vertex is greater than n/2
(ii) The sum of degrees of any non-adjacent pair of
vertices is greater than n
(iii) 1 (n2 – 3n 6) 1 (25 – 15 6) 8.
2
2
Thus the condition m t 1 (n2 3n 6) is satisfied.
2
(iv) The sum of degrees of each pair of vertices in
the given graph is greater than n 1 5 1 4.
Thus four sufficiency conditions are satisfied
(whereas one condition out of these four conditions
is sufficient for the existence of Hamiltonian path/
circuit). Hence the graph has a Hamiltonian circuit.
For example, the following circuits in G are
Hamiltonian (Figure 2.88).
a
b
and
e
c
d
a
b
Solution.
Here
Number of vertices (n) 5
Number of edges (m) 4
deg(a) deg(b) deg(c) deg(d) 1
deg(e) 4
We note that
5
2
(ii) deg(a) deg(b) 2 ≥/ 5, that is sum of any nonadjacent pair of vertices is not greater than 5
(iii) 1 (n2–3n 6) 1 (25 –15 6) 8
2
2
1
Therefore the condition m t (n2 – 3n 6) is not
2
satisfied,
(i) deg(a) deg(b) deg(c) deg(d) ≥/
(iv) deg(a) deg(b) 2 ≥/ 4, i.e., the condition that
sum of degrees of each pair of vertices in the
graph is not greater than or equal to n 1.
Hence no sufficiency condition is satisfied. So we
try the Proposition 2.1. Suppose that G has a Hamiltonian circuit. Then G should have a subgraph
which contains every vertex of G, and number of
vertices and number of edges in H should be same.
Thus, H should have five vertices a, b, c, d, e and
five edges. Since G has only four edges, H cannot
have more than four edges. Hence no such subgraph
is possible. Hence, the given graph does not have
Hamiltonian circuit.
e
c
EXAMPLE 2.36
Does the graph shown below (Figure 2.90) possess a
Hamiltonian circuit?
d
Figure 2.88
EXAMPLE 2.35
Does the graph shown below (Figure 2.89) has
Hamiltonian circuit?
a
b
e
c
d
Figure 2.89
A
D
B
C
Figure 2.90
Solution. In the given graph
Number of vertices (n) 4
Number of edges (m) 8
Degree of each vertex 4

Graphs
Thus we see that
(i) Degree of each vertex is greater than n/2
(ii) Sum of degree of each pair of vertices is
greater than n 1,
1
1
(16 – 12 6) 5 and so m t
(iii) 1 (n2 – 3n 6)
2
2
2
2
(n – 3n 6).
Hence the given graph has a Hamiltonian path. For
example, ABCDA is a Hamiltonian path in the given
graph (see Figure 2.91).
A
D
B
C
Figure 2.91
EXAMPLE 2.37
Is the graph shown in Figure 2.92 a Hamiltonian?
B
A
E
2.25
(iii) deg(A) deg(E) 2 2 4 and so the condition
that “the sum of degrees of non-adjacent vertices is greater than or equal to n” is not satisfied
(iv) “The sum of degrees of any pair of vertices is
greater than or equal to n 1” is not satisfied.
So suppose that G has a Hamiltonian circuit. Then it
should have a connected subgraph H containing six
vertices, six edges and degree of each vertex should be
2. To have degree of each vertex equal to 2, we should
remove two edges from C and two edges from B.
For example, now, if we remove CE from C,
then deg(E) z 2, If we remove CF, then deg(F) z 2.
So, we cannot remove CE and CF. If we remove CD,
then deg(D) z 2. If we remove CA, then deg(A) z 2.
Hence no such subgraph H exists. So, G cannot have
a Hamiltonian circuit.
Remark 2.6. Since the degree of each vertex in the
above graph is even, it has Eulerian circuit. Thus
the graph in EXAMPLE 2.37 is Eulerian but not
Hamiltonian.
EXAMPLE 2.38
Is the graph given in Figure 2.93 Hamiltonian?
A
B
C
D
E
F
F
D
C
Figure 2.92
Solution. We note that
deg(A) deg(D) deg(E) deg(F) 2,
deg(B) deg(C) 4.
Further,
Number of vertices (n) 6
Number of edges (m) 8.
So,
1 (n2 – 3n 6) 1 (36 – 18 6) 12.
2
2
Thus the conditions
(i) Degree of each vertex is greater than or equal
to n/2 is not satisfied
(ii) m t 1 (n2 – 3n 6) is not satisfied
2
Figure 2.93
Solution. In this graph,
Number of vertices (n) 6
Number of edges (m) 9.
So,
1 (n2 – 3n 6) 1 (36 – 18 6) 12,
2
2
deg(A) deg(C) 2,
deg(D) deg(F) 3 deg(E),
deg(B) 5.
Thus, no sufficient condition is satisfied in this case.
But, the graph has Hamiltonian circuit ADEFCBA
shown in the Figure 2.94.
2.26

Engineering Mathematics-II
A
B
C
D
E
F
For example, if vertices in a graph denote recreational sites of a town and weights of edges denote
the distances in kilometers between the sites, then
the graph shown in Figure 2.96 is a weighted graph.
7
D
Figure 2.94
7
EXAMPLE 2.39
Show that the graph G shown in Figure 2.95 is not
Hamiltonian.
a
e
d
g
c
b
f
h
i
Figure 2.95
Solution. If G is Hamiltonian, then it has a subgraph
H that
1.
2.
3.
4.
Contains every vertex of G
Is connected
Has the same number of edges as vertices
Is such that every vertex has degree 2
Thus, if such a graph exists, then it will have vertices
(a, b, c, d, e, f, g, h, i), will be connected, will have six
edges and the degree of each vertex shall be 2. We
note that deg(d ) deg(e) deg(f ) 2. The degree of
the vertex h in G is 3, one edge incident on h must be
deleted from G to create H. The edge {h, e} is required
since otherwise deg(e) z 2. So we have to delete either
{g, h} or {h, i}. If we delete {h, i} and retain {g, h},
then {g, i} has to be deleted to keep g of degree 2. In
such a case, there is only one edge {i, f } on i and so
there is a contradiction to (4). So we cannot remove
{h, i}. Similarly, we see that {g, h} cannot be deleted.
Hence no such subgraph H exists and so G does
not have a Hamiltonian circuit.
Definition 2.55. A weighted graph is a graph for which
each edge or each vertex or both is (are) labelled with
a numerical value, called its weight.
A
C
14
11
5
9
B
Figure 2.96
Definition 2.56. The weight of an edge (vi, vj) is called
distance between the vertices vi and vj.
Definition 2.57. A vertex u is a nearest neighbour
of vertex v in a graph if u and v are adjacent and
no other vertex is joined to v by an edge of lesser
weight than (u, v).
For example, in the above example, B is the
nearest neighbour of C, whereas A and C are both
nearest neighbours of the vertex D. Thus, nearest
neighbour of a set of vertices is not unique.
Definition 2.58. A vertex u is a nearest neighbour of a
set of vertices {v1, v2, …, vn} in a graph if u is adjacent to some member vi of the set and no other vertex
adjacent to member of the set is joined by an edge of
lesser weight than (u, vi).
In the above example, if we have set of vertices
as {B, D}, C is the neatest neighbour of {B, D}
because the edge (C, B) has weight 5 and no other
vertex adjacent to {B, D} is linked by an edge of
lesser weight than (C, B).
Definition 2.59. The length of a path in a graph is the
sum of lengths of edges in the path.
Definition 2.60. Let G (V, E) be a graph and let lij
denote the length of edge (vi, vj) in G. Then a shortest path from vi to vk is a path such that the sum
l12 l23 lk 1,k of lengths of its edges is minimum, that is, total edge weight is minimum.
2.7.1
Travelling Salesperson Problem
This problem requires the determination of a shortest Hamiltonian circuit in a given graph of cities

Graphs
and lines of transportation to minimize the total fare
for a travelling person who wants to make a tour of
n cities visiting each city exactly once before returning home.
The weighted graph model for this problem
consists of vertices representing cities and edges
with weight as distances (fares) between the cities. The salesman starts and ends his journey at
the same city and visits each of n 1 cities once
and only once. We want to find minimum total
distance.
We discuss the case of five cities and so consider
the weighted graph shown in the Figure 2.97.
2.27
(i) Let us choose a as the starting vertex. Then d
is the nearest vertex and then (a, d) is the corresponding edge. Thus we have the figure
a
7
c
b
e
d
(ii) From d, the nearest vertex is c, so we have a
path shown below:
a
a
12
14
7
7
9
b
c
b
6
6
5
13
c
10
8
e
d
d
11
e
Figure 2.97
(iii) From c, the nearest vertex is e. So we have the
path as shown below:
a
We shall use nearest neighbour algorithm to solve
the problem:
Algorithm: Nearest neighbour (closest insertion)
Input: A weighted complete graph G.
Output: A sequence of labelled vertices that forms
a Hamiltonian cycle.
Start at any vertex v.
Initialize l(v) 0.
Initialize i 0.
7
c
b
6
8
e
d
(iv) From e, the nearest vertex is b and so we have
the path
a
While there are unlabelled vertices
i: i 1.
7
c
b
Traverse the cheapest edge that join v to an unlabelled vertex, say w
Set l(w) i.
v: w.
For the present example,
5
d
6
8
e
(v) Now, from b, the only vertex to be covered is a
to form Hamiltonian circuit. Thus we have a
2.28

Engineering Mathematics-II
Hamiltonian circuit as given below. The length
of this Hamiltonian circuit is
2 3 2 5 4 5 4 6 31.
EXAMPLE 2.41
Find a Hamiltonian circuit of minimal weight for the
graph shown below (Figure 2.100)
7 6 8 5 14 40.
a
14
7
6
5
10
8
However, this is not Hamiltonian circuit of minimal length.
The total distance of a minimum Hamiltonian
circuit (Figure 2.98) is 37.
a
7
10
c
6
5
8
9
d
12
Figure 2.100
Solution. Starting from the point a and using nearest
neighbour method, we have the required Hamiltonian circuit as a b c d a with total length as
10 10 8 12 40.
9
b
c
15
a
e
d
10
b
c
b
Definition 2.61. A k-factor of a graph is a spanning
subgraph of the graph with the degree of its vertices being k.
Consider the graph shown in Figure 2.101.
e
d
Figure 2.98
a
Total length 7 6 9 5 10 37.
Remark 2.7. Unless otherwise stated, try to start
from a vertex of largest weight.
EXAMPLE 2.40
Find a Hamiltonian circuit of minimal weight for the
graph given below (Figure 2.99)
b
2
d
3
3
4
a
2
c
5
4
e
2
6
c
d
e
f
Figure 2.101
f
6
5
b
g
5
Then the graph shown in Figure 2.102 shows a
1-factor of the given graph.
4
h
Figure 2.99
Solution. Starting from c and applying nearest neighbour method, we have the required Hamiltonian
circuit as c a b d e f g h c with total length as
Figure 2.102
Graphs

2.29
Solution. The adjacency matrix A(G) (aij) is the
matrix such that
Also then, (see Figure 2.103)
aij Number of edges connecting vi and vj.
So, for the given graph, the adjacency matrix is
Figure 2.103
is a 2-factor of the given graph.
2.8
MATRIX REPRESENTATION OF GRAPHS
A graph can be represented inside a computer by
using the adjacency matrix or the incidence matrix
of the graph.
Definition 2.62. Let G be a graph with n ordered vertices v1, v2, …, vn. Then the adjacency matrix of G
is the n u n matrix A(G) (aij) over the set of nonnegative integers such that
⎡0
⎢1
⎢
A(G ) = ⎢1
⎢
⎢1
⎢⎣1
EXAMPLE 2.42
Find the adjacency matrix of the graph shown below
(Figure 2.104):
v1
v3
v2
v4
Figure 2.104
v5
1
1
0
0
0
1
1
0
0
0
1⎤
1⎥
⎥
0⎥ .
⎥
0⎥
0 ⎥⎦
EXAMPLE 2.43
Find the adjacency matrix of the graph given below:
v2
v1
aij the number of edges connecting vi and vj for all
i, j 1, 2, …, n.
We note that if G has no loop, then there is no
edge joining vi to vi, i 1, 2, …, n.Therefore, in this
case, all the entries on the main diagonal will be 0.
Further, if G has no parallel edge, then the entries
of A(G) are either 0 or 1.
It may be noted that adjacent matrix of a graph
is symmetric.
Conversely, given a n u n symmetric matrix
A(G) (aij) over the set of non-negative integers, we
can associate with it a graph G, whose adjacency
matrix is A(G), by letting G have n vertices and joining vi to vertex vj by aij edges.
1
0
1
1
1
v4
v3
Figure 2.105
Solution. We note that there is a loop at v1 and parallel edges between v1, v3, and v3, v4. So the adjacency
matrix A(G) is given by the following 4 u 4 matrix:
⎡1
⎢0
A(G) ⎢
⎢2
⎢
⎣0
0
0
1
0
2
1
0
2
0⎤
0⎥
⎥.
2⎥
⎥
0⎦
EXAMPLE 2.44
Find the graph that has the following adjacency
matrix:
⎡1
⎢2
⎢
⎢1
⎢
⎣2
2
0
2
1
1
2
1
0
2⎤
1⎥
⎥.
0⎥
⎥
0⎦
2.30

Engineering Mathematics-II
Solution. We note that there is a loop at v1 and a loop
at v3. There are parallel edges between v1, v2; v1, v4;
v2, v1; v2, v3; v3, v2; v4, v1. Thus the graph is as shown
in Figure 2.106.
v1
v2
v3
v4
2
1
0
0
3
1
2
0
0
3
0⎤
0⎥
⎥
3⎥
⎥
3⎥
0 ⎥⎦
⎡2
⎢1
⎢
A4 = ⎢0
⎢
⎢0
⎢⎣ 3
1
2
0
0
3
0
0
5
4
0
0
0
4
5
0
3⎤
3⎥
⎥
0⎥ .
⎥
0⎥
6 ⎥⎦
⎡3
⎢1
⎢
B = A + A2 + A3 + A4 = ⎢ 3
⎢
⎢1
⎢⎣ 4
The following theorem is stated without proof.
Theorem 2.14. Let G be a graph with n vertices v1, v2,
…, vn and let A(G) denote the matrix of G. If B (bij)
is the matrix
B A A2 An 1,
then the graph G is a connected graph if and only
if bij z 0 for i z j, that is, B has no zero entry off the
main diagonal.
EXAMPLE 2.45
A graph G has the following adjacency matrix. Verify whether it is connected.
0
0
0
1
0
1
0
0
0
1
0
1
0
0
1
0⎤
0⎥
⎥
1⎥ .
⎥
1⎥
0 ⎥⎦
0
1
0
0
1
0
0
2
1
0
0
0
1
2
0
1⎤
1⎥
⎥
0⎥ ,
⎥
0⎥
2⎥⎦
1
3
1
3
4
3
1
7
5
4
1
3
5
7
4
4⎤
4⎥
⎥
4⎥ .
⎥
4⎥
8 ⎥⎦
Since B has no zero entry off the main diagonal, the
graph is connected.
Definition 2.63. Suppose a graph G has n vertices v1,
v2, …, vn and t edges e1, e2, …, et. The incidence
matrix B(G) of G is the n u t matrix B(G) (bij),
where
bij the number of times that the vertex vi is incident with the edge ej,
that is,
⎧0 if vi is not end of e j
⎪
bij = ⎨1 if vi is an end of the non-loop e j
⎪2 if v is an end of the loop e
i
j
⎩
Solution. We have
⎡1
⎢0
⎢
A2 = ⎢0
⎢
⎢0
⎢⎣1
0
0
1
2
0
Therefore,
Figure 2.106
⎡0
⎢0
⎢
A = ⎢1
⎢
⎢0
⎢⎣0
⎡0
⎢0
⎢
A3 = ⎢ 2
⎢
⎢1
⎢⎣0
2.9
PLANAR GRAPHS
Definition 2.64. A graph which can be drawn in the
plane so that its edges do not cross is said to be planar.
For example, the graph shown in Figure 2.107 is
planar:
Graphs
A
2.31
But the map K5 is not planar because in this case, the
edges cross each others (see Figure 2.110).
B
C
D
E
B
A
E
Figure 2.107
Also the complete graph K4 shown below (Figure
2.108) is planar.
A
C
D
Figure 2.110
B
D
C
Figure 2.108
In fact, it can be redrawn as shown below in Figure
2.109 so that no edges cross.
A
Definition 2.65. An area of the plane that is bounded
by edges of the planar graph and is not further subdivided into subareas is called a region or face of
the planar graph.
A face is characterized by the cycle that forms
its boundary.
Definition 2.66. A region is said to be finite if its area
is finite and infinite if its area is infinite. Clearly a
planar graph has exactly one infinite region.
For example, consider the graphs shown in Figure 2.111.
B
D
C
Figure 2.109
(i)
(ii)
1
2
G21
G1
1
3
2
4
6
5
5
6
7
9

4
8
10
Figure 2.111
G2
3
2.32

Engineering Mathematics-II
We note that in the graph G1, there are five regions A, B, C, D, E as shown below:
1
4
3
5
B
C
6
A
7
3
8
4
Finite region
Finite region
Finite region
1
E
3
3
4
9
8
D
9
Infinite region
10
Infinite region
8
Infinite region
In graph G2, there are four region A, B, C, D (Figure 28.112)
1
2
2
1
5
B
C
A
6
3
4
4
Finite region
Finite region
1
E
Finite region
2
D
5
3
4
Infinite region
Figure 2.112
6

Graphs
Definition 2.67. Let f be a face (region) in a planar
graph. The length of the cycle (or closed walk)
which borders f is called the degree of the region f.
It is denoted by deg(f ).
In a planar graph we note that each edge either
borders two regions or is contained in a region
and will occur twice in any walk along the border
of the region. Thus we have the following:
Theorem 2.15. The sum of the degrees of the regions
of a map is equal to twice the number of edges.
For example, in the graph G2, discussed above,
we have
deg(A) 4, deg(B) 3, deg(C) 4, deg(d) 5.
The sum of degrees of all regions 4 3 4 5 16.
2.33
a
x
G
Figure 2.113
We delete “a” and the edge x incident on “a” from
the graph G. The resulting graph Gc (Figure 2.114)
has n edges and so by induction hypothesis, the formula holds for Gc. Since G has one more edge than
Gc, one more vertex than Gc and the same number
of faces as Gc, it follows that the formula v – e r 2
holds also for G.
Number of edges in G2 8.
Hence the sum of degrees of regions is twice the
number of edges.
Theorem 2.16. (Euler’s Formula for Connected Planar
Graphs) If G is a connected planar graph with e
edges, v vertices and r regions, then
v – e r 2.
Proof: We shall use induction on the number of
edges. Suppose that e 0. Then the graph G consists
of a single vertex, say P. Thus, G is as shown below:
Figure 2.114
Now suppose that G contains a cycle. Let x be an
edge in a cycle as shown in Figure 2.115.
x
•P
and we have e 0, v 1, r 1. Thus 1 – 0 1 2 and
the formula holds in this case.
Suppose that e 1. Then the graph G is one of the
two graphs shown below:
,
e = 1, v = 2, r = 1
G′
e = 1, v = 1, r = 2
We see that, in either case, the formula holds.
Suppose that the formula holds for connected
planar graph with n edges. We shall prove that this
holds for graph with n 1 edges. So, let G be the
graph with n 1 edges. Suppose first that G contains
no cycles. Choose a vertex v1 and trace a path starting at v1. Ultimately, we will reach a vertex a with
degree 1, that we cannot leave (see Figure 2.113).
G
Figure 2.115
Now the edge x is part of a boundary for two faces.
We delete the edge x but no vertices to obtain the
graph Gc as shown in Figure 2.116.
G′
Figure 2.116
2.34

Engineering Mathematics-II
Thus Gc has n edges and so by induction hypothesis the formula holds. Since G has one more face
(region) than Gc, one more edge than Gc and the
same number of vertices as Gc, it follows that the
formula v – e r 2 also holds for G. Hence, by
mathematical induction, the theorem is true.
Remark 2.8. Planarity of a graph is not affected if
(i) An edge is divided into two edges by the insertion of new vertex of degree 2 (Figure 2.117).
EXAMPLE 2.46
Show that the graph K3,3, given in Figure 2.120
below, is not planar.
c1
c2
c3
c4
c5
c6
K3,3
Figure 2.120
Figure 2.117
(ii) Two edges that are incident with a vertex of
degree 2 are combined as a single edge by the
removal of that vertex (Figure 2.118).
Figure 2.118
Definition 2.68. Two graphs G1 and G2 are said to
be isomorphic to within vertices of degree 2 (or
homeomorphic) if they are isomorphic or if they can
be transformed into isomorphic graphs by repeated
insertion and/or removal of vertices of degree 2.
Definition 2.69. The repeated insertion/removal of
vertices of degree 2 is called sequence of series
reduction.
For example, the graphs shown in Figure 2.119
are isomorphic to within vertices of degree 2.
A problem based on this example can be stated
as “Three cities c1, c2 and c3 are to be directly connected by express ways to each of three cities c4,
c5 and c6. Can this road system be designed so that
the express ways do not cross?” This example shows
that it cannot be done.
Solution. Suppose that K3,3 is planar. Since every cycle
in K3,3 has at least four edges, each face (region) is
bounded by at least four edges. Thus the number of
edges that bound regions is at least 4r. Also, in a planar graph each edge belongs to at most two bounding
cycles. Therefore,
2e t 4r (sums of degrees of region is equal to twice
the number of edges)
But, by Euler’s formula for planar graph,
r e – v 2.
Hence,
2e t 4(e – v 2).
(1)
In case of K3,3 we have e 9, v 6 and so (1) yields
18 t 4 (9 – 6 2) 20,
which is a contradiction. Therefore K3,3 is not planar.
and
Remark 2.9. By a argument similar to the above
example, we can show that the graph K5 (Figure
2.121) is not planar.
Figure 2.119
If we define a relation R on the set of graphs by
G1 R G2 if G1 and G2 are homeomorphic, then R is
an equivalence relation. Each equivalence class consists of a set of mutually homeomorphic graphs.
(Non-planar graph K5)
Figure 2.121
Graphs
We observe that if a graph contains K3,3 or K5 as a
subgraph, then it cannot be planar.
The following theorem, which we state without
proof, gives necessary and sufficient condition for a
graph to be planar.
2.35
we can then carry out series reduction. The resulting
graph will have nine edges.
Also we know that K3,3 has nine edges. So this
approach seems promising. Using trial and error, we
find that the edge (g, h) should be removed. Then g
and h have degree 2.
Kuratowski’s Theorem 2.17. A graph G is planar if and
only if G does not contain a subgraph homeomorphic to K3,3 or K5.
The complete graph K5 and the complete bipartite graph K3,3 are called the Kuratowski graphs.
a
f
b
g
EXAMPLE 2.47
Using Kuratowski’s theorem show that the graph G,
shown in Figure 2.122, is not planar.
h
e
c
d
(Graph obtained by deleting edges (a, b) and (f, e)).
a
f

b
g
a
h
e
f
b
g
c
d
h
G
e
Figure 2.122
c
d
Solution. Let us try to find K3,3 in the graph G. We
know that in K3,3, each vertex has degree 3. But we
note that in G, the degree of a, b, f and e each is 4.
So we eliminate the edges (a, b) and (f, e) so that
all vertices have degree 3. If we eliminate one more
edge, we will obtain two vertices of degree 2 and
(Graph obtained by eliminating the edge (g, h).
Performing series reduction now, we obtain an isomorphic copy of K3,3 (Figure 2.123).
a
f
b
e
c
d
(Isomorphic copy of K3,3, obtained by series reduction)
Figure 2.123
2.36

Engineering Mathematics-II
v – e r 2.
Hence, by Kurtowski’s theorem, the given graph G
is not planar.
EXAMPLE 2.48
If the graph given in Figure 2.124 is planar, redraw
it so that no edges cross, otherwise find a subgraph
homeomorphic to either K5 or K3,3.
a
b
f
c
e
d
So,
Hence,
6d3 v–e
e
.
3
v
e d 3v – 6
or
Remark 2.10. Using above theorem, let us consider
planarity of K5. Here v 5, e 10. Suppose K5 is planar, then by the above theorem,
6 d 3(5) – 10 5,
which is impossible. Hence K5 is non-planar.
Definition 2.70. Any graph homeomorphic either to K5
or to K3,3 is called Kuratowski subgraph.
Figure 2.124
Solution. The given graph can be redrawn as shown
in the Figure 2.125.
EXAMPLE 2.49
Find Kuratowski subgraph of the following graph
(Figure 2.126):
b
a
2e
3
2 v–erdv–e
f
a
b
e
f
g
h
c
e
d
d
c
Figure 2.126
Figure 2.125
We then observe that no edges cross. Hence the
given graph is planar.
Theorem 2.18. Let G be a connected planar graph
with v vertices, e edges, where v t 3. Then e d 3v – 6
or 6 d 3v – e.
(Note that the theorem is not true for K1, where
v 1 and e 0 and is not true for K2, where v 2 and
e 1).
Proof: Let r be the number of regions in a planar representation of the graph G. The sum of the degrees
of the regions is equal to 2e. But each region has
degree greater than or equal to 3. Hence 2 e t 3 r, that
is, r d 2e . But, by Euler’s formula,
3
Solution. In the given graph,
Number of vertices 8
Number of edges 12.
Firstly, we remove the edge (g, h). Then deg(g) 2,
deg(h) 2 and so the vertices g and h can be removed.
Then, we have
Number of vertices 6
Number of edges 9
and the graph reduces to
a
b
e
d
f
,
c

Graphs
which is homeomorphic to K3,3 (Kuratowski graph)
shown in Figure 2.127.
c
a
This subgraph is homeomorphic to K3,3 as shown in
Figure 2.130.
f
e
K3,3
d
a
e
c
f
b
K3,3
d
b
Figure 2.127
Thus the given graph is non-planar by Kuratowski
theorem.
EXAMPLE 2.50
Find a subgraph homeomorphic to either K5 or K3,3
in the graph given below (Figure 2.128):
a
e2
e12
e7
f
b
e1
e13
e9
e10 e6
c
e8
e3
e4
e
2.37
e11
e5
Figure 2.130
This also shows, by Kuratowski theorem, that the
given graph is not planar.
2.10
COLOURING OF GRAPH
Definition 2.71. Let G be a graph. The assignment of
colours to the vertices of G, one colour to each vertex, so that the adjacent vertices are assigned different colours is called vertex colouring or colouring
of the graph G.
d
Definition 2.72. A graph G is n-colourable if there
exists a colouring of G which uses n colours.
Figure 2.128
Solution.
Here,
Definition 2.73. The minimum number of colours
required to paint (colour) a graph G is called
the chromatic number of G and is denoted by
c (G).
Number of vertices 6
Number of edges 13
deg(a) 5, deg( f ) 4, deg(c) 4, deg(d) 5
But, in K3,3, the degree of each vertex is 3. We delete
the edges (a, c), ( f, d), (a, e), (b, d), then degree of
each vertex a, f, c and d becomes 3. As such, the subgraph becomes as given below (Figure 2.129), which
has six vertices each with degree 3 and has 9 edges.
a
EXAMPLE 2.51
Find the chromatic number for the graph shown in
the Figure 2.131.
b
b
a
f
c
e
d
Figure 2.129
c
d
e
G
Figure 2.131
2.38

Engineering Mathematics-II
Solution. The triangle a b c needs three colours. Suppose that we assign colours c1, c2, c3 to a, b and c
respectively. Since d is adjacent to a and c, d will
have different colour than c1 and c3. So we paint d
by c2. Then e must be painted with a colour different
from those of a, d and c, that is, we cannot colour e
with c1, c2 or c3. Hence, we have to give e a fourth
colour c4. Hence
c(G) 4.
Welsh–Powell algorithm to determine upper
bound to the chromatic number of a given graph.
The input is a given graph G.
Solution. We note that
deg( f ) 5,
deg(a) deg(b) deg(c) deg(d) deg(e) 3.
Step 1. Ordering the vertices according to decreasing degree yields
f, a, b, c, d, e
Step 2.
Step 3.
Step 4.
Step 5.
Paint f with colour c1.
Paint a, d with colour c2.
Paint b, e with colour c3.
Paint c with colour c4.
1. Order the vertices of G according to decreasing
degree.
2. Assign the first colour, say c1, to the first vertex
and then, in sequential order, assign c1 to each
vertex, which is not adjacent to a previous vertex
assigned c1.
3. Repeat Step 2 with a second colour c2 and the
subsequence of the remaining non-painted vertices.
4. Repeat Step 3 with a third colour c3, then a
fourth colour c4 and so on until all vertices are
coloured.
5. Exit.
Hence c(G) d 4. Also since there is a triangle
in the given graph so we have c(G) t 3. Thus
3 d c(G) d 4. But we do not yet know exactly
what the chromatic number is. We try to build a
3-colouring of G.
Let us start colouring the triangle abf with the
colours c1, c2, c3, respectively. Since c is adjacent to
the vertices b and f of colour c2 and c3, respectively, c
is forced to be coloured c1 and then d is forced to be
c2. However, now the adjacent vertices a and e cannot both have colour c1. Thus the graph cannot be
3-coloured. But using a fourth colour c4 for e gives
us 4-colouring of G. Hence, c(G) 4.
EXAMPLE 2.52
Use Welsh-Powell algorithm to determine an upper
bound to the chromatic number of the “Wheel” graph
shown in Figure 2.132.
EXAMPLE 2.53
Use Welsh-Powell algorithm to colour the graph
shown in Figure 2.133.
a
b
a
e
f
e
d
d
c
Figure 2.132
c
b
f
h
g
Figure 2.133
Solution. Ordering the vertices according to decreasing degrees, we get the sequence e, c, g, a, b, d, f, h.
Use the colour c1 to colour (paint) e and a.
Use the colour c2 to paint c, d and h.
Use third colour c3 to paint vertices g, b and f.
c2
Some rules for colouring a graph
1. c(G) ≤ | V |, where | V | is the number of vertices
of G.
2. A triangle always requires three colours, that is,
c(K ) = 3. Similarly c(K ) = n, where K is the
3
n
n
complete graph of n vertices.
3. If some subgraph of G requires k-colours, then
c(G) ≥ k.
4. If deg(v) = n, then at most n colours are required
to colour the vertices adjacent to v.
5. c(G) = max {c(C): C is a connected component
of G}.
6. Every n-colourable graph has at least n vertices
v such that deg(v) ≥ n − 1.
7. c(G) ≤ 1 + Δ(G), where Δ(G) is the largest degree
of any vertex of G.
8. The following statements are equivalent:
(i) A graph G is 2-colourable
(ii) G is bipartite
(iii) Every cycle of G has even length
9. If d (G) is the smallest degree of any vertex of
G, then
χ(G ) ≥
|v|
.
| v | − δ(G )
EXAMPLE 2.54
Find chromatic number of the graphs shown in
Figure 2.134.
c1
c1
c2
c2
c3
and
c1
Thus all the vertices are painted such that no adjacent vertices get the same colour. Hence c(G) = 3.
2.39

Graphs
c2
c1
c2
c1
(i)
(ii)
Figure 2.134
Solution. The graph (i) is 2-colourable whereas the
graph (ii) is 3-colourable.
Theorem 2.19. Bipartite graph is 2-colourable unless
G is edgeless.
Proof: A two colouring is obtained by assigning one
colour to every vertex in one of the bipartition parts
and another colour to every vertex in the other partition part as shown in Figure 2.135.
c1
c2
c1
c2
c2
Figure 2.135
Corollary 2.3. Even cycle graphs C2n have c(C2n) = 2.
Proof: An even cycle graph is bipartite and therefore, by Theorem 2.19, its chromatic number is 2.
For example consider the graph shown in
Figure 2.136.
a
b
c
d
Figure 2.136
2.40

Engineering Mathematics-II
It is equivalent to the following bipartite graph
shown in Figure 2.137.
K1
a
c
b
d
Figure 2.137
Therefore, it is 2-colourable.
Proposition 2.2. Odd cycle graph C2n + 1 has
c(C ) = 3.
2n + 1
Proof: Let v1, v2, …, v2n, v2n + 1 be vertices of a cycle
graph c2n + 1. If two colours were sufficient, then they
would have to alternate around the cycle. Thus, the
odd subscripted would have to be one colour and the
even subscripted vertices have to be second colour
but vertex v2n + 1 is adjacent to v1, and so according to
this scheme two adjacent vertices v1 and v2n + 1 have
the same colour. This is a contradiction and so C2n + 1
is not 2-colourable but 3-colourable.
It follows, therefore, that “The chromatic number of a cycle is either two or three, depending on
whether its length is even or odd.”
Definition 2.74. The join G + H of the graphs G and H
is obtained from the graph G ∪ H by adding an edge
between each vertex of H.
For example, if we have a graph as shown below
(Figure 2.138):
Figure 2.139
Regarding the chromatic number of the join G + H
of the graphs G and H we have the following:
Proposition 2.3. The join of a graph G and H has
chromatic number
c(G + H) = c(G) + c(H).
Definition 2.75. The n vertex wheel graph Wn is called
an odd-order wheel if n is odd and an even-order
wheel if n is even.
Proposition 2.4. Odd-order wheel graph has
c(W
) = 3, for all m ≥ 1.
2m + 1
Proof: Using the fact that the wheel graph W2m + 1 is
the join C2m + K1, it follows that
c(W
) = c(C2m) + c(K1) = 2 + 1 = 3.
2m + 1
K1
Figure 2.140 (W5)
Proposition 2.5. Even-order wheel graph has c(W2m) = 4.
Proof: Using the fact that W2m = c2m −1 + K1, it follows that
X(W2m) = c(c2m − 1) + c(K1) = 3 + 1 = 4.
(G)
Figure 2.138
K1
and a graph
∑ K 1,
then the join G + K1 of the graphs G and K1 is the
graph as shown below (Figure 2.139):
Figure 2.141 (W6)
Graphs
EXAMPLE 2.55
Find chromatic number of the graph shown in
Figure 2.142:
a

2.41
Definition 2.78. Let G be a directed graph. The indegree of a vertex v of G is the number of edges
ending at v. It is denoted by indeg(v).
EXAMPLE 2.56
Consider the directed graph shown below (Figure 2.143):
d
v1
b
e1
c
Figure 2.142
v2
Solution. Here degree of each vertex is 3. Therefore,
c(G) ≤ 1 + 3 = 4. Since it has a triangle, c(G) ≥ 3.
Thus, 3 ≤ c(G) ≤ 4.
We first colour the triangle abd with colours c1, c2,
c3, respectively. Since c is adjacent to each of a, b and
d, therefore a different colour c4 have to be given to it.
Hence the graph is 4-colourable and so c(G) = 4.
4
= 4.
Note: If we apply rule 9, then χ ≥
(4 − 3)
Hence χ(G ) = 4.
e2
Theorem 2.20 (Four-Colour Theorem of Appel
and Haken)
Any planar graph is 4-colourable.
(Proof of the theorem is out of the scope of this
book).
2.11
DIRECTED GRAPHS
Definition 2.76. A directed graph or digraph consists of two finite sets:
(i) A set V of vertices (or nodes or points).
(ii) A set E of directed edges (or arcs), where each
edge is associated with an ordered pair (v, w) of
vertices called its endpoints. If edge e is associated with the ordered pair (v, w), then e is said
to be directed edge from v to w.
The directed edges are indicated by arrows.
We say that edge e = (v, w) is incident from v and
is incident into w.
The vertex v is called initial vertex and the vertex w
is called the terminal vertex of the directed edge (v, w).
v3
e4
e5
v5
v4
e6
Figure 2.143
Here edge e1 is (v2, v1) whereas e6 is denoted by
(v5, v5) and is called a loop. The indegree of v2 is 1,
outdegree of v2 is 3.
Definition 2.79. A vertex with 0 indegree is called a
source, whereas a vertex with 0 outdegree is called
a sink.
For instance, in the above example, v1 is a sink.
Definition 2.80. If the edges and/or vertices of a
directed graph G are labelled with some type of data,
then G is called a labelled directed graph.
Definition 2.81. Let G be a directed graph with
ordered vertices v1, v2, …, vn. The adjacency matrix
of G is the matrix A = (aij) over the set of non-negative integers such that
aij = the number of arrows from vi to vj, i, j = 1, 2, …, n.
EXAMPLE 2.57
Find the adjacency matrices for the graphs given
below (Figure 2.144):
v1
b
v2
c
a
,
d
Definition 2.77. Let G be a directed graph. The outdegree of a vertex v of G is the number of edges
beginning at v. It is denoted by outdeg(v).
e3
v3
v4
(i)
(ii)
Figure 2.144
2.42

Engineering Mathematics-II
Solution. The edges in the directed graph are (a, a),
(b, b), (c, c), (d, d), (c, a), (c, b) and (d, b). Therefore
the adjacency matrix A = (aij) is
⎡1
⎢0
⎢
⎢1
⎢
⎣0
0
1
1
1
0⎤
0⎥
⎥.
0⎥
⎥
1⎦
0
0
1
0
Figure 2.146
A directed graph which is not simple is called
directed multi-graph.
(iii) The edges in the graph in (ii) are (v2, v3), (v1,
v1), (v1, v3), (v3, v1), (v3, v4), (v4, v3). Hence the
adjacency matrix is
⎡1
⎢0
⎢
⎢1
⎢
⎣0
0
0
0
0
0⎤
0⎥
⎥.
1⎥
⎥
0⎦
1
1
0
1
Definition 2.83. A simple digraph is said to be
strongly connected if for each pair of vertices v, w,
there are path, from v to w and w to v.
For example, the graphs show in Figure 2.147 are
strongly connected.
v
w
EXAMPLE 2.58
Find the directed graph represented by the adjacency matrix:
⎡0
⎢0
⎢
⎢0
⎢
⎢1
⎢⎣0
1
0
0
1
0
0
1
0
0
0
0
0
1
0
0
0⎤
0⎥
⎥
1⎥ .
⎥
0⎥
0 ⎥⎦
Figure 2.147
Solution. We observe that a12 = 1, a23 = 1, a34 = 1, a35 = 1,
a41 = 1, a42 = 1. Hence the digraph is as shown in Figure 2.145.
v1
v2
v4
v3
v5
Figure 2.145
Definition 2.82. In a directed graph, if there is no
more than one directed edge in a particular direction
between a pair of vertices, then it is called simple
directed graph.
For example, the graph shown in Figure 2.146 is
a simple directed graph.
Definition 2.84. Let G be a simple digraph with n vertices. Then a n × n matrix P = (pij) such that
⎧1 if there is a path from vi to v j
pij = ⎨
⎩0 otherwise
is called path matrix or reachability matrix of G.
The path matrix can be given in term of adjacency matrix:
P = A ∨ A(2) ∨ A(3) ∨ … ∨ A(n)
and
A(2) = A 䉺 A, A(r) = A(r − 1) 䉺 A,
where ∨ represents Boolean addition and 䉺 denotes
the Boolean matrix multiplication.
Warshall’s algorithm for finding path matrix
from adjacency matrix
We discuss this algorithm with the help of the
following example.
Graphs
EXAMPLE 2.59
Find the adjacency matrix and the path matrix for
the digraph shown in Figure 2.148.
b
c
a
d
Figure 2.148
Solution. The adjacency matrix of this digraph is:
⎡0
⎢1
A= ⎢
⎢0
⎢
⎣0
1
0
0
0
0
1
0
0
0⎤
0⎥
⎥.
1⎥
⎥
0⎦
Then the path matrix P = (pij) is found as follows:
We take
⎡0
⎢1
P0 = ⎢
⎢0
⎢
⎣0
1
0
0
0
0⎤
0⎥
⎥ = A itself.
1⎥
⎥
0⎦
0
1
0
0
Now we find P1 by consulting column 1 and row 1.
We note that P0 has 1 in location 2 of column 1 and
location 2 of row 1. Thus the element p22 is replaced
by 1 in P0. Thus we have
⎡0
⎢1
P1 = ⎢
⎢0
⎢
⎣0
1
1
0
0
0
1
0
0
0⎤
0⎥
⎥.
1⎥
⎥
0⎦
To find P2 we consult column 2 and row 2 of P1. We
note that P1 has 1 in locations 1 and 2 of column 2
and locations 1, 2 and 3 of row 2. Thus, to obtain P2
we should put 1 in positions p11, p12, p13, p21, p22 and
p23 of matrix P1. We thus have
⎡1
⎢1
P2 = ⎢
⎢0
⎢
⎣0
1
1
0
0
1
1
0
0
0⎤
0⎥
⎥.
1⎥
⎥
0⎦

2.43
We now consult column 3 and row 3 of P2. The
matrix P2 has 1 in locations 1 and 2 of column 3 and
1 in location 4 of row 3. Thus we shall obtain P3 by
putting 1 in positions p14, p24. Thus
⎡1
⎢1
P3 = ⎢
⎢0
⎢
⎣0
1
1
0
0
1
1
0
0
1⎤
1⎥
⎥.
1⎥
⎥
0⎦
Finally, we consult column 4 and row 4 of P3. We
have 1 in location 1, 2, 3 of column 4 and there is
no 1 in row 4. Hence P4 = P3. Thus the path matrix is
⎡1
⎢1
P = ( pij ) = ⎢
⎢0
⎢
⎣0
1 1 1⎤
1 1 1⎥
⎥.
0 0 1⎥
⎥
0 0 0⎦
Remark 2.11. Given an adjacency matrix of a graph
G, we first draw digraph of G and then apply definition of path matrix. We will obtain path matrix.
Now, we state a result without proof.
Theorem 2.21. Let A be the adjacency matrix of a graph
G with n vertices and let Bn = A + A2 + A3 + … + An.
Then the path matrix P and Bn have the same nonzero entries.
Let G be a strongly connected directed graph.
Then for any pair of vertices v and w in G, there is a
path from v to w and from w to v. Accordingly, G is
strongly connected if and only if the path matrix
P of G has no zero entries.
Thus, in view of the above theorem, we have:
Theorem 2.22. Let A be adjacency matrix of a graph
G with n vertices and let Bn = A + A2 + … + An. Then
G is strongly connected if and only if Bn has no zero
entries.
EXAMPLE 2.60
A directed graph has the following adjacency
matrix. Check whether it is strongly connected.
⎡0 1 1⎤
A(G ) = ⎢⎢1 0 1 ⎥⎥ .
⎣⎢1 1 0 ⎥⎦
2.44

Engineering Mathematics-II
Solution. To find the path matrix, we first consider
column 1 and row 1 of
⎡0 1 1⎤
P0 = ⎢⎢1 0 1 ⎥⎥ = A (G ).
⎢⎣1 1 0 ⎥⎦
In column 1, the entry 1 is located at 2 and 3 positions
and 1 is at locations 2 and 3 in row 1. Thus put 1 at
p22, p23, p32, p33 and get
Figure 2.149
Remark 2.12. In the above example, we also note that
⎡0 1 1⎤
P1 = ⎢⎢1 1 1⎥⎥ .
⎢⎣1 1 1⎥⎦
B3 = A + A2 + A3
⎡ 0 1 1 ⎤ ⎡ 2 1 1 ⎤ ⎡ 2 3 3⎤
= ⎢⎢1 0 1 ⎥⎥ + ⎢⎢1 2 1 ⎥⎥ + ⎢⎢ 3 2 3⎥⎥
⎢⎣1 1 0 ⎥⎦ ⎢⎣1 1 2⎥⎦ ⎢⎣ 3 3 2⎥⎦
⎡4 5 5⎤
= ⎢⎢ 5 4 5 ⎥⎥ (all non-zero entries)
⎣⎢ 5 5 4 ⎦⎥
We note that 1 is located in column 2 of P1 at position
1, 2, 3, and 1 is located at position 1, 2, 3 of row 2. Thus
put 1 at p11, p12, p13, p21, p22, p23, p31, p32, p33, and get
⎡1 1 1⎤
P2 = ⎢⎢1 1 1⎥⎥ .
⎢⎣1 1 1⎥⎦
showing that G is strongly connected.
The next step will not change the position. Hence,
⎡1 1 1⎤
P = ⎢⎢1 1 1⎥⎥ (all non-zero entries).
⎢⎣1 1 1⎥⎦
Therefore, G is strongly connected.
The graph of G is shown in Figure 2.149.
(i) Trivial tree
2.12
TREES
Definition 2.85. A graph is said to be a tree if it is a
connected acyclic graph.
A trivial tree is a graph that consists of a single
vertex. An empty tree is a tree that does not have
any vertices or edges.
For example, the graphs shown in Figure 2.150
are all trees.
(ii) Tree of 3 vertices
(iv) Tree of 13 vertices
(iii) Tree of 4 vertices
(v) Tree of 8 vertices
Figure 2.150

Graphs
But the graphs shown in Figure 2.151 are not trees:
Proof: Let T be a particular but arbitrary chosen tree
having more than one vertex (Figure 2.153).
e
(i) Has a cycle and
so is not a tree
2.45
v
e′
v′
e′′
(ii) Has a cycle and
so is not a tree
T
v′′
Figure 2.153
(iii) Disconnected graph
and so is not a tree
Figure 2.151
Definition 2.86. A collection of disjoint trees is called
a forest.
Thus a graph is a forest if and only if it is circuit
free.
Definition 2.87. A vertex of degree 1 in a tree is called
a leaf or a terminal node or a terminal vertex.
Definition 2.88. A vertex of degree greater than 1 in
a tree is called a branch node or internal node or
internal vertex.
Consider the tree shown in Figure 2.152.
b
c
f
a
e
d
g
h
i
Figure 2.152
In this tree, the vertices b, c, d, f, g, i are leaves
whereas the vertices a, e, h are branch nodes.
2.12.1
Characterization of Trees
We have the following interesting characterization
of trees:
Lemma 2.1. A tree that has more than one vertex has
at least one vertex of degree 1.
1. Choose a vertex v of T. Since T is connected and
has at least two vertices, v is not isolated and
there is an edge e incident on v.
2. If deg(v) > 1, there is an edge e′ ≠ e because there
are, in such a case, at least two edges incident on
v. Let v′ be the vertex at the other end of e′. This
is possible because e′ is not a loop by the definition of a tree.
3. If deg(v′) > 1, then there are at least two edges
incident on v′. Let e″ be the other edge different
from e′ and v″ be the vertex at other end of e″.
This is again possible because T is acyclic.
4. If deg(v″) > 1, repeat the above process. Since the
number of vertices of a tree is finite and T is circuit free, the process must terminate and we shall
arrive at a vertex of degree 1.
Remark 2.13. In the proof of the Lemma 2.1, after
finding a vertex of degree 1, if we return to v and
move along a path outward from v starting with e,
we shall reach to a vertex of degree 1 again. Thus
it follows that “Any tree that has more than one
vertex has at least two vertices of degree 1”.
Lemma 2.2. There is a unique path between every two
vertices in a tree.
Proof: Suppose on the contrary that there are more
than one path between any two vertices in a given
tree T. Then T has a cycle which contradicts the
definition of a tree because T is acyclic. Hence the
lemma is proved.
Lemma 2.3. The number of vertices is one more than
the number of edges in a tree.
Equivalently, we may state this lemma as
2.46

Engineering Mathematics-II
“For any positive integer n, a tree with n vertices
has n − 1 edges”.
Proof: We shall prove the lemma by mathematical
induction.
Let T be a tree with one vertex. Then T has no
edges, that is, T has 0 edge. But 0 = 1–1. Hence the
lemma is true for n = 1.
Suppose that the lemma is true for k > 1. We
shall show that it is then true also for k + 1. Since the
lemma is true for k, the tree has k vertices and k − 1
edges. Let T be a tree with k + 1 vertices. Since k is
+ve, k + 1 ≥ 2 and so T has more than one vertex. Hence,
by Lemma 2.1, T has a vertex v of degree 1. Also there
is another vertex w and so there is an edge e connecting v and w. Define a subgraph T ′ of T so that
V(T ′) = V(T) – {v},
E(T ′) = E(T) – {e}.
Then number of vertices in T ′ = (k + 1) – 1 = k
and since T is circuit free and T ′ has been obtained
on removing one edge and one vertex, it follows
that T ′ is acyclic. Also T ′ is connected. Hence T ′
is a tree having k vertices and therefore by induction hypothesis, the number of edges in T ′ is k − 1.
But then
Lemma 2.4. A graph in which there is a unique path
between every pair of vertices is a tree
(This lemma is converse of Lemma 2.2).
Proof: Since there is a path between every pair of
points, therefore the graph is connected. Since a
path between every pair of points is unique, there
does not exist any circuit because existence of circuit implies existence of distinct paths between pair
of vertices. Thus the graph is connected and acyclic
and so is a tree.
Lemma 2.5. (Converse of Lemma 2.3)
A connected graph G with e = v – 1 is a tree
First Proof: The given graph is connected and
e = v – 1. To prove that G is a tree, it is sufficient
to show that G is acyclic. Suppose on the contrary
that G has a cycle. Let m be the number of vertices in this cycle. Also, we know that number of
edges in a cycle is equal to number of vertices
in that cycle. Therefore number of edges in the
present case is m. Since the graph is connected,
every vertex of the graph which is not in cycle
must be connected to the vertices in the cycle (see
Figure 2.154).
Number of edges in T = number of edges in T ′ + 1
= k – 1 + 1 = k.
Thus the Lemma is true for tree having k + 1 vertices. Hence the lemma is true by mathematical
induction.
Corollary 2.4. Let C(G) denote the number of components of a graph. Then a forest G on n vertices has
n − C(G) edges.
Proof: Apply Lemma 2.3 to each component of the
forest G.
Corollary 2.5. Any graph G on n vertices has at least
n – C(G) edges.
Proof: If G has cycle-edges, remove them one at a
time until the resulting graph G* is acyclic. Then G*
has n – C(G*) edges by Corollary 2.4. Since we have
removed only circuit, C(G*) = C(G). Thus G* has
n – C(G) edges. Hence G has at least n – C(G) edges.
Figure 2.154
Now each edge of the graph that is not in the cycle can
connect only one vertex to the vertices in the cycle.
There are v − m vertices that are not in the cycle.
So the graph must contain at least v − m edges that
are not in the cycle. Thus we have e ≥ v – m + m = v,
which is a contradiction to our hypothesis. Hence
there is no cycle and so the graph is a tree.
Second Proof: We shall show that a connected graph
with v vertices and v – 1 edges is a tree. It is sufficient
to show that G is acyclic. Suppose on the contrary
Graphs
that G is not circuit free and has a nontrivial circuit
C. If we remove one edge of C from the graph G,
we obtain a graph G′ which is connected (see Figure 2.155).
2.47

Since
e = e′ + e″ + …,
v = v′ + v″ + …,
we have
e < v – 1,
which contradicts our hypotheses. Hence G is
connected. So G is connected and acyclic and is
therefore a tree.
EXAMPLE 2.61
Construct a graph that has six vertices and five
edges but is not a tree.
C
G
G′
Figure 2.155
If G′ still has a nontrivial circuit, we repeat the
above process and remove one edge of that circuit
obtaining a new connected graph. Continuing this
process, we obtain a connected graph G* which is
circuit free. Hence G* is a tree. Since no vertex has
been removed, the tree G* has v vertices. Therefore,
by Lemma 2.3, G* has v − 1 edges. But at least one
edge of G has been removed to form G*. This means
that G* has not more than v – 1 – 1 = v – 2 edges.
Thus, we arrive at a contradiction. Hence our supposition is wrong and G has no cycle. Therefore G is
connected and cycle free and so is a tree.
Lemma 2.6. A graph G with e = v – 1, that has no circuit is a tree.
Proof: It is sufficient to show that G is connected.
Suppose G is not connected and let G′, G″, … be the
connected components of G. Since each of G′, G″,
… is connected and has no cycle, they all are tree.
Therefore, by Lemma 2.3,
e′ = v′ − 1,
e″ = v″ − 1,
————
————,
where e′, e″, … are the number of edges and v′,
v″, … are the number of vertices in G′, G″, …,
respectively. We have, on adding
e′ + e″ + … = (v′ − 1) + (v″ − 1) + …
Solution. We have
Number of vertices = 6
Number of edges = 5
So the condition e = v – 1 is satisfied. Therefore, to
construct graph with six vertices and five edges that
is not a tree, we should keep in mind that the graph
should not be connected. The graph shown in Figure
2.156 has six vertices and five edges but is not connected. Hence it is not a tree.
v4
v1
v2
v3
v5
v6
Figure 2.156
Definition 2.89. A directed graph is said to be a
directed tree if it becomes a tree when the direction
of edges are ignored.
For example, the graph shown in Figure 2.157 is a
directed tree.
Figure 2.157
Definition 2.90. A directed tree is called a rooted
tree if there is exactly one vertex whose incoming
degree is 0 and the incoming degrees of all other
vertices are 1.
2.48

Engineering Mathematics-II
The vertex with incoming degree 0 is called the
root of the rooted tree.
A tree T with root v0 will be denoted by (T, v0).
Definition 2.91. In a rooted tree, a vertex, whose outgoing degree is 0 is called a leaf or terminal node,
whereas a vertex whose outgoing degree is non-zero
is called a branch node or an internal node.
Definition 2.92. Let u be a branch node in a rooted
tree. Then a vertex v is said to be child (son or offspring) of u if there is an edge from u to v. In this
case, u is called parent (father) of v.
Definition 2.97. Let u be a branch node in the tree
T = (V, E). Then the subgraph T ′ = (V ′, E ′) of T
such that the vertices set V ′ contains u and all of
its descendents and E′ contains all the edges in all
directed paths emerging from u is called a subtree
with u as the root.
Definition 2.98. Let u be a branch node. By a subtree
of u, we mean a subtree that has child of u as root.
In the above example, we note that the Figure
2.159 is a subtree of T,
u
Definition 2.93. Two vertices in a rooted tree are said
to be siblings (brothers) if they are both children of
same parent.
Definition 2.94. A vertex v is said to be a descendent
of a vertex u if there is a unique directed path from
u to v.
In this case, u is called the ancestor of v.
Definition 2.95. The level (or path length) of a vertex
u in a rooted tree is the number of edges along the
unique path between u and the root.
w
s
t
Figure 2.159
whereas the Figure 2.160 is a subtree of the branch
node u.
w
Definition 2.96. The height of a rooted tree is the
maximum level to any vertex of the tree.
As an example of these terms, consider the rooted
tree shown in Figure 2.158:
Root ………..………....…… Level 0
,
v
s
t
Figure 2.160
EXAMPLE 2.62
Let
V = {v1, v2, v3, v4, v5, v6, v7, v8}
x …………… u ………………...…… Level 1
and let
E = {(v2, v1), (v2, v3), (v4, v2), (v4, v5), (v4, v6), (v6, v7),
(v5, v8)}.
y …………. z… v………….
w ………..…… Level 2
Show that (V, E) is rooted tree. Identify the root of
this tree.
t …………
s ……… Level 3
Figure 2.158
Here y is a child of x; x is the parent of y and z. Thus
y and z are siblings. The descendents of u are v, w, t
and s. Levels of vertices are shown in the figure. The
height of this rooted tree is 3.
Solution. We note that
Incoming degree of v1 = 1,
Incoming degree of v2 = 1,
Incoming degree of v3 = 1,
Incoming degree of v4 = 0,
Incoming degree of v5 = 1,
Incoming degree of v6 = 1,
Graphs
Incoming degree of v7 = 1,
Incoming degree of v8 = 1.
Since incoming degree of the vertex v4 is 0, it follows that v4 is root.
Further,
Outgoing degree of v1 = 0,
Outgoing degree of v3 = 0,
Outgoing degree of v7 = 0,
Outgoing degree of v8 = 0.
Therefore, v1, v3, v7, v8 are leaves. Also,
Outgoing degree of v2 = 2,
Outgoing degree of v4 = 3,
Outgoing degree of v5 = 1,
Outgoing degree of v6 = 1.
Now the root v4 is connected to v2, v5 and v6. So, we
have the following diagram:
v4 (Root)
v5
v2
(i) Since v1 is not mother of itself, (v1, v1) ∉T.
(ii) If (v1, v2) ∈ T, then v1 is mother of v2. But then v2
cannot be mother of v1. Hence (u1, v2) ∈ T does
not imply (v2, v1) ∈ T.
(iii) If (v1, v2) ∈ T, (v2, v3) ∈ T, then
v1 is mother of v2
v2
v1
and
v2 is mother of v3.
This does not imply that v1 is mother of v3, i.e., (v1,
v3) ∉ T. Thus T is irreflexive, asymmetric and is not
transitive. Hence T is not an equivalence relation.
Consider the rooted trees T and T ′ shown in the
Figure 2.162, where T is family tree of a man who
has two sons, with the elder son having no child and
younger having three children.
v6
v4 (Root)
v8
2.49
Solution. The graph of T is cycle free, because
descendents cannot be parents of their parents. The
graph is also connected. Hence T is rooted tree. We
further note that
Now v2 is connected to v1 and v3, v5 is connected to
v8, v6 is connected to v7. Thus, we have the graph
shown in Figure 2.161.
v5

v6
v3
v7
Figure 2.161
We thus have a connected acyclic graph and so (V,
E) is a rooted tree with root v4.
EXAMPLE 2.63
Let F be the set of all female descendent of a lady v0. We
define a relation T on F as follows: if v1, v2, ∈ F, then
v1 T v2 if v1 is mother of v2. Show that the relation T
on F is a rooted tree with root v0. Is this relation an
equivalence relation?
T′
T
Figure 2.162
Although T ′(as a graph) is isomorphic to T, it could
be the family tree of another man whose elder son
has three children and whose younger son has no
child.
The above discussion motivates the following
definitions:
Definition 2.99. A rooted tree in which the edges incident from each branch node are labelled with integers 1, 2, 3, … is called an ordered tree.
2.13
ISOMORPHISM OF TREES
Definition 2.100. Two ordered trees are said to be
isomorphic if (i) there exists a one-to-one correspondence between their vertices and edges and that
preserves the incident relation (ii) labels of the corresponding edges match.
In view of this definition, the ordered trees shown
in Figure 2.163 are not isomorphic.
2.50

Engineering Mathematics-II
2
1
1
2
1
3
2
Solution. Let f be a function defined by
2
1
f (z) = v,
3
f (s) = u,
Figure 2.163
EXAMPLE 2.64
Show that the tree T1 and T2 shown in the diagram
below are isomorphic.
1
c
a
Then f is an one-one, onto mapping which preserves
adjacency. Hence, T1 and T2 are isomorphic.
Remark 2.14. There are three non-isomorphic trees
(Figure 2.166) with five vertices:
3
d
a
4
T1
T2
x
b
5
l
u
In the tree T2,
deg(3) = 4.
Further, deg(a) = deg(1) = 1, deg(c) = deg(2), deg(d) =
deg(4) = deg(e) = 1 = deg(5) and deg(b) = deg(3) = 4.
Thus we may define a function f from the vertices of
T1 to the vertices of T2 by
f (d) = 4,
f (e) = 5.
This is a one-to-one and onto function. Also, adjacency relation is preserved because if vi and vj are
adjacent vertices in T1, then f (vi) and f (vj) are
adjacent vertices in T2. Hence T1 is isomorphic
to T2.
EXAMPLE 2.65
Show that the tree T1 and T2, shown in the Figure
2.165 are isomorphic
y
z
s
u
v
l
T2
T1
Figure 2.165
t
r
v
Definition 2.101. Let T1 and T2 be rooted trees with
roots r1 and r2, respectively. Then T1 and T2 are
isomorphic if there exists a one-one, onto function
f from the vertex set of T1 to the vertex set of T2
such that
(i) Vertices vi and vj are adjacent in T1 if and only
if the vertices f (vi) and f (vj) are adjacent in T2,
(ii) f (r1) = r2.
The function f is then called an isomorphism.
EXAMPLE 2.66
Show that the trees T1 and T2 shown in Figure 2.167
are isomorphic.
w1
v1
w
v2
t
s
Figure 2.166
deg(b) = 4.
x
z
w
Solution. We observe that in the tree T1,
f (c) = 2,
y
e
Figure 2.164
f (b) = 3,
f (t) = l.
c
e
f (a) = 1,
f (x) = m,
2
d
b
f (y) = w,
v5
m
v3
v6
v4
v7
w2
w3
w5
w6
w8
v8
Figure 2.167
w4
w7

Graphs
2.51
Solution. We observe that T1 and T2 are rooted trees.
Define f: (vertex set of T1) → (vertex set of T2) by
f (v1) = w1,
f (v2) = w3,
f (v3) = w4
f (v4) = w2,
f (v5) = w6,
f (v6) = w7
f (v7) = w5,
f (v8) = w8.
Then f is one-to-one and adjacency relation is preserved. Hence f is an isomorphism and so the rooted
trees T1 and T2 are isomorphic
Figure 2.169
whereas the tree shown in Figure 2.170 is a fully
binary tree.
EXAMPLE 2.67
Show that the rooted trees shown in Figure 2.168 are
not isomorphic:
a
1
Figure 2.170
c
b
e
f
g
d
h
2
3
5
7
6
T1
4
T2
Figure 2.168
Solution. We observe that the degree of root in T1 is
3, whereas the degree of root in T2 is 4. Hence T1 is
not isomorphic to T2.
Definition 2.102. An ordered tree in which every
branch node has at most n offspring is called a nary tree (or n-tree).
Definition 2.103. An n-ary tree is said to be fully
n-ary tree (complete n-ary tree or regular n-ary
tree) if every branch node has exactly n offspring.
Definition 2.106. Let T1 and T2 be binary trees with
roots r1 and r2, respectively. Then T1 and T2 are isomorphic if there is a one-one, onto function f from
the vertex set of T1 to the vertex set of T2 satisfying
(i) Vertices vi and vj are adjacent in T1 if and only
if the vertices f (vi) and f (vj) are adjacent in T2
(ii) f (r1) = r2
(iii) v is a left child of w in T1 if and only if f (v) is a
left child of f (w) in T2
(iv) v is a right child of w in T1 if and only if f (v) is
a right child of f (w) in T2
The function f is then called an isomorphism
between binary trees T1 and T2.
EXAMPLE 2.68
Show that the trees given in Figure 2.171 are isomorphic.
v1
Definition 2.104. An ordered tree in which every
branch node has at most two offspring is called a
binary tree (or 2-tree).
v2
v3
Definition 2.105. A binary tree in which every branch
node (internal vertex) has exactly two offspring is
called a fully binary tree.
For example, the tree given in Figure 2.169 is a
binary tree,
w1
and
w2
w3
v4
w4
Figure 2.171
Solution. Define f by f (vi) = wi, i = 1, 2, 3, 4. Then f
satisfies all the properties for isomorphism. Hence
T1 and T2 are isomorphic.
2.52

Engineering Mathematics-II
EXAMPLE 2.69
Show that the trees given in Figure 2.172 are not
isomorphic.
Root
v
s
w1
v1
v2
w
w2
,
v3
w3
Figure 2.173
v4
w5
v5
T1
w4
are, respectively the trees shown in Figure 2.174.
T2
Figure 2.172
s
w
Solution. Since the root v1 in T1 has a left child but the
root w1 in T2 has no left child, the binary trees are not
isomorphic.
and
(Left subtree of v)
Definition 2.107. Let v be a branch node of a binary
tree T. The left subtree of v is the binary tree whose
root is the left child of v, whose vertices consists of
the left child of v and all its descendents and whose
edges consists of all those edges of T that connects
the vertices of the left subtree together.
The right subtree can be defined analogously.
For example, the left subtree and the right subtree
of v in the tree (Figure 2.173)
+
Figure 2.174
2.14
REPRESENTATION OF ALGEBRAIC
EXPRESSIONS BY BINARY TREES
Binary trees are used in computer science to represent algebraic expressions involving parentheses.
For example, the binary trees (Figure 2.175) represent the expressions, a + b, a/b and b * (c * d),
respectively.
/
*
,
a
b
(Right subtree of v)
and
a
b
b
*
c
d
Figure 2.175
Thus, the central operator acts as root of the tree.
EXAMPLE 2.70
Draw a binary tree to represent
(i) (2 – (3 × x)) + ((x – 3) – (2 + x))
(ii) ab – (c/(d + e)).
Solution
(i) In this expression + is the central operator.
Therefore the root of tree is + . The binary tree
is shown in Figure 2.176.
Graphs
2.53
We note that there shall be three rounds to declare
the champion. Let
+
−
i = Number of games played
= number of internal vertices
p = Number of leaves = number of players.
−
2

×
+
−
Then, we see that
x
3
x
3
i = p – 1 = 7.
x
2
In general, for complete n-ary tree, we will have
Figure 2.176
(ii) Here the central operator is −. Therefore it is the
root of the tree. We have the following binary
tree (Figure 2.177) to represent this expression:
−
c
b
To derive this formula we first prove the following
result:
Theorem 2.23. If T is a full binary tree with i internal
vertices, then T has i + 1 terminal vertices (leaves)
and 2i + 1 total vertices.
/
a
(n – 1) i = p – 1.
+
e
d
Figure 2.177
EXAMPLE 2.71
Consider a single elimination tournament with eight
players in a wrestling game. How many rounds will
be there to declare the champion and how many
games are to be played?
Solution. In this example, the leaves of the tree shown
in Figure 2.178 will represent player, the branch
nodes will represent winners in the first round and
the second round. Then the tree showing the tournament shall be as given below: The graph of single
elimination tournament is a full binary tree. The
champion shall be at the root.
Proof: The vertices of T consist of the vertices that
are children (of some parent) and the vertices that
are not children (of any parent). There is a non-child
(the root). Since there are i internal vertices, each
having two children, there are 2i children. Thus the
total number of vertices of T is 2i + 1 and the number of terminal vertices is
(2i + 1) – i = i + 1.
This completes the Proof.
In the context of above example, we have
Number of leaves = p = i + 1
⇒ i = p – 1.
Remark 2.15. In case of full n-ary tree, if i denotes the
number of branch nodes, then total number of vertices
of T is ni + 1 and the number of terminal vertices is
ni + 1 – i = i(n − 1) + 1.
If p is the number of terminal vertices, then
p = i(n – 1) + 1,
Champion
which yields
(n – 1) i = p – 1.
….... Second round winners ……..
…………..
1
2
3
First round winners
4
5
Figure 2.178
……….…..
6
7
8
EXAMPLE 2.72
Find the minimum number of extension cords, each
having four outlets, required to connect 22 bulbs to
a single electric outlet.
Solution. Clearly, the graph of the problem is a regular quaternary tree with 22 leaves.
2.54

Engineering Mathematics-II
Let i denote the internal vertices and p denote
the number of leaves, then using the expression
(n – 1)i = p – 1, we have
(a) Suppose that the root of T has only one child
(Figure 2.180).
Root
21
(4 − 1) i = 22 − 1 or i = = 7.
3
Thus seven extension cords, as shown in Figure 2.179,
are required.
Figure 2.180
14
13
1 2
3
4
15 16
5 6 7
17
8 9
18
10 11
19 20
12
21 22
Eliminating the root and the edge incident on the
root, we get a tree T ′ of height h – 1 in which number
of leaves is same as in T. Using induction hypothesis
on T ′, we get p ≤ 2h − 1. Since 2h −1 < 2h, we have p ≤ 2h
and the inequality holds in this case.
(b) Suppose the root of T has offspring’s v and w
(Figure 2.181).
Figure 2.179
Root
EXAMPLE 2.73
A computing machine has been given instruction
which computes the sum of three numbers. How
many times the addition instruction will be executed
to perform the sum of 11 numbers?
Solution. In this problem, we will have a regular
ternary tree with 11 leaves. Thus n = 3, p = 11 and
so the relation (n – 1)i = p – 1 yields
i=
p − 1 10
=
= 5.
n −1 2
Theorem 2.24. If a binary tree of height h has p leaves,
then
log2 p ≤ h or p ≤ 2h.
Proof: It is sufficient to prove that
p ≤ 2h,
(1)
because then taking logarithm to the base 2 of both
sides, we will get
log2 p ≤ h.
We shall prove the inequality (1) by induction on h.
If h = 0, then the binary tree consists of a single
vertex and p = 1 and so (1) is satisfied.
Now suppose that (1) holds for a binary tree whose
height is less than h. We shall prove that it holds for
a tree of height h. So let T be a tree with height h > 0
with p leaves. We consider the following two cases:
v
w
Figure 2.181
Let T1 be the subtree rooted at v whose height is h1
and has p1 leaves. Similarly, let T2 be the subtree
rooted at w whose height is h2 and has p2 leaves.
Using mathematical induction, to these trees, we
h
h
have p1 ≤ 2 1 and p2 ≤ 2 2 . But the leaves of T consist of the leaves of T1 and T2. Hence,
p = p1 + p2 ≤ 2h1 + 2h2
≤ 2h − 1 + 2h − 1 = 2h − 1 (1 + 1) = 2h.
Hence, by mathematical induction, the result holds.
It can be proved on the same lines that an n-ary
tree of height h has atmost nh leaves.
EXAMPLE 2.74
Find the number of terminal vertices in a regular
binary tree of height 4.
Solution. Let p be the number of terminal vertices.
We are given that height (h) of the tree is 4. So, we
must have p ≤ 24 = 16. Since the tree is regular, we
have p = 16.

Graphs
Root
2.55
EXAMPLE 2.77
How many binary trees are possible on three vertices?
Solution. We know that the number of different trees
on n vertices is given by
bn =
(A regular binary tree of height 4 with 16 leaves)
So we have
Figure 2.182
EXAMPLE 2.75
Does there exist a full binary tree with 12 internal
vertices and 15 leaves?
bn =
Solution. No, since
p ≤ 2h = 26 = 64.
We now state some results without proof on n-ary tree.
Theorem 2.25. The number bn of different binary
trees on n vertices is given by
bn =
1 ⎛ 2n⎞
.
n + 1 ⎜⎝ n ⎟⎠
Theorem 2.26. A regular n-ary tree of height h has at
least n + (n − 1) (h − 1) leaves.
1 ⎛ 6⎞ 1 ⎛ 6 × 5 × 4 × 3 × 2 × 1⎞
= ⎜
⎟ = 5 trees
3 + 1 ⎜⎝ 3⎟⎠ 4 ⎝ 3 × 2 × 1 × 3 × 2 × 1 ⎠
and those are shown in Figure 2.183.
Solution. We know that if i is the number of branch
nodes in a full binary tree, then the number of leaves
is i + 1. Therefore for a tree with 12 branch nodes,
the number of leaves should be 13 and not 15. Hence
such tree does not exist.
EXAMPLE 2.76
Is there a binary tree with height 6 and 65 leaves?
1 ⎛ 2n⎞
.
n + 1 ⎜⎝ n ⎟⎠
,
,
,
,
Figure 2.183
2.15
SPANNING TREE OF A GRAPH
Definition 2.108. Let G be a graph, then a subgraph of
G which is a tree is called tree of the graph.
Definition 2.109. A spanning tree for a graph G is a
subgraph of G that contains every vertex of G and
is a tree.
Thus,
“A spanning tree for a graph G is a spanning
subgraph of G which is a tree”.
EXAMPLE 2.78
Determine a tree and a spanning tree for the connected graph given below (Figure 2.184):
Theorem 2.27. If in a rooted regular n-ary tree,
I = Sum of the path length of all branch nodes
E = Sum of the path length of all leaves
i = Number of branch nodes
then
E = (n – 1)I + n i
Thus, in particular, for a full binary tree,
E = I + 2i.
G
Figure 2.184
Solution. The given graph G contains circuits and we
know that removal of the circuits gives a tree. So,
we note that the Figure 2.185 is a tree.
2.56

Engineering Mathematics-II
Definition 2.111. A chord (or a link) of a tree is an
edge of the graph that is not in the tree.
It follows from the above remark that the number
of chords in a tree is equal to m – n + 1, where n is the
number of vertices and m is the number of edges in
the graph related to the tree.
Figure 2.185
And the Figure 2.186 is a spanning tree of the graph G.
EXAMPLE 2.80
Consider the graph discussed in the above example.
We note that the edge (v2, v3) is a branch of the tree
T1, whereas (v1, v3) is a chord of the tree T1.
Figure 2.186
EXAMPLE 2.79
Find all spanning trees for the graph G shown below
(Figure 2.187).
v2
v3
v1
v6
v4
v5
Figure 2.187
Solution. The given graph G has a circuit v1 v2 v3 v1. We
know that removal of any edge of the circuit gives a tree.
So the spanning trees of G are shown in Figure 2.188.
v2
v3
v2
v6
v3
v6
,
v1
v4
,
v1
v5
v4
T1
v5
T2
v2
v1
v3
v4
Definition 2.112. The set of the chords of a tree is
called the complement of the tree.
v6
v5
Figure 2.188
Remark 2.16. We know that a tree with n vertices has
exactly n – 1 edges. Therefore if G is a connected graph
with n vertices and m edges, a spanning tree of G must
have n – 1 edges. Hence the number of edges that must
be removed before a spanning tree is obtained must be
m – (n – 1) = m – n + 1.
For illustration, in EXAMPLE 2.79, n = 6, m = 6, so,
we had to remove one edge to obtain a spanning tree.
Definition 2.110. A branch of a tree is an edge of the
graph that is in the tree.
Theorem 2.28. A graph G has a spanning tree if and
only if G is connected.
Proof: Suppose first that a graph G has a spanning
tree T. If v and w are vertices of G, then they are also
vertices in T and since T is a tree there is a path from v
to w in T. This path is also a path in G. Thus every two
vertices are connected in G. Hence G is connected.
Conversely, suppose that G is connected. If G is
acyclic, then G is its own spanning tree and we are
done. So suppose that G contains a cycle C1. If we
remove an edge from the cycle, the subgraph of G so
obtained is also connected. If it is acyclic, then it is a
spanning tree and we are done. If not, it has at least
one circuit, say C2. Removing one edge from C2, we
get a subgraph of G which is connected. Continuing
in this way, we obtain a connected circuit free subgraph T of G. Since T contains all vertices of G, it is
a spanning tree of G.
Theorem 2.29. (Cayley’s Formula). The number of spanning trees of the complete graph Kn, n ≥ 2 is nn − 2.
(Proof of this formula is out of scope of this book)
EXAMPLE 2.81
Find all the spanning trees of K4 (Figure 2.189).
Solution. According to Cayley’s formula, K4 has
44 − 2 = 42 = 16 different spanning trees.
v4
v1
v3
K4
Figure 2.189
v2
Graphs
Here n = 4, so the number of edges in any tree
should be n – 1 = 4 – 1 = 3. But here number of edges
is equal to 6. So to get a tree, we have to remove
three edges of K4. The 16 spanning trees so obtained
are shown in Figure 2.190.

2.57
We now discuss algorithm due to E · W. Dijkstra
which efficiently solves the shortest path problem.
The idea is to grow a Disjkstra tree, starting at the
vertex s, by adding, at each iteration, a frontier
edge, whose non-tree endpoint is as close as possible to s. The algorithm involves assigning labels
to vertices.
For each tree vertex x, let dist [x] denote the
distance from vertex s to x and for each edge e in
the given weighted graph G, let w(e) be its edgeweight.
After each iteration, the vertices in the Dijkstra
tree (the labelled vertices) are those to which the
shortest paths from s have been found.
Priority of the Frontier Edges: Let e be a frontier edge and let its P value be given by
v4
v3
v4
v3
v1
v2
v1
v2
v4
v3
v4
v3
v1
v2
v1
v2
v4
v3
v4
v3
v1
v2
v1
v2
v4
v3
v4
v3
v1
v2
v1
v2
v4
v3
v4
v3
v1
v2
v1
v2
v4
v3
v4
v3
We are now in a position to describe Dijkstra’s shortest path algorithm.
v1
v2
v1
v2
2.16.1 Dijkstra’s Shortest Path Algorithm
v4
v3
v4
v3
v1
v2
v1
v2
v4
v3
v4
v3
v1
v2
v1
v2
Input: A connected weighted graph G with nonnegative edge-weights and a vertex s of G.
Output: A spanning tree T of G, rooted at the vertex s, whose path from s to each vertex v is a shortest
path from s to v in G and a vertex labelling giving
the distance from s to each vertex.
Initialize the Dijkstra tree T as vertex s.
Initialize the set of frontier edges for the tree T
as empty.
P(e) = dist [x] + w(e),
Figure 2.190
2.16
SHORTEST PATH PROBLEM
Let s and t be two vertices of a connected weighted
graph G. Shortest path problem is to find a path
from s to t whose total edge weight is minimum.
where x is the labelled endpoint of e and w(e) is the
edge-weight of e. Then,
(i) The edge with the smallest P value is given the
highest priority.
(ii) The P value of this highest priority edge e gives
the distant from the vertex s to the unlabelled
endpoint of e.
dist: [s] = 0.
Write label 0 on vertex s.
While Dijkstra tree T does not yet span G.
For each frontier edge e for T,
2.58

Engineering Mathematics-II
Let x be the labelled endpoint of edge e.
Let y be the unlabelled endpoint of edge e.
Set P(e) = dist [x] + w(e)
Let e be a frontier edge for T that has small
est P value
Let x be the labelled endpoint of edge e
Let y be the unlabelled endpoint of edge e
Add edge e (and vertex y) to tree T
s
13
v
16
w
6
y
14
5
x
If t is the labelled endpoint of edge e, then P values
are given by
P(e) = dist [t] + w(e),
Return Dijkstra tree T and its vertex labels.
EXAMPLE 2.82
Apply Dijkstra algorithm to find shortest path from s
to each other vertex in the graph given in Figure 2.191.
where dist [t] = distance from s to t and w(e) is the
edge weight of edge e. For each vertex v, dist [v]
appears in the parenthesis. Iteration tree at the end of
each iteration is drawn in dark line.
Iteration 2
Iteration 1
s(0)
s(0)
8
10
13
v
z(8)
8
10
z(8)
11
11
w
17
7
Figure 2.191
Write label dist [y] on vertex y.
13
z
11
dist [y]: P(e)
v
8
10
16
6
17
7
w(13)
16
6
y
14
y
14
5
7
17
5
x
x
dist [s] = 0
P(sw) = 13 (minimum)
dist [z] = 8
P(zy) = 8 + 7 = 15
dist [s] = 0
P(zy) = 8 + 7 =15 (minimum)
P(sy) = 16
dist [z] = 8
P(zx) = 8 + 17 = 25
P(zv) = 8 + 10 = 18
dist [w] = 13
P(zv) = 8 + 10 = 18
P(zw) = 8 + 11 = 19
P(sy) = 16
P(zx) = 8 + 17 = 25
P(wx) = 13 + 14 = 27
Iteration 3
Iteration 4
s(0)
s(0)
v
13
8
10
z(8)
13
8
10
z(8)
11
11
16
w(13)
v(18)
6
17
14
16
w(13)
6
y(15)
5
x
7
17
14
y(15)
5
x
7

Graphs
dist [s] = 0
P(zv) = 18 (minimum)
Dist [s] = 0
P(yx) = 20 (minimum)
dist [z] = 8
P(zx) = 8 + 17 = 25
Dist [z] = 8
P(zx) = 8 + 17 = 25
dist [w]= 13
P(wx) = 13 + 14 = 27
Dist [w] = 13
P(vx) = 18 + 6 = 24
dist [y] = 15
P(yx) = 15 + 5 = 20
Dist [y] = 15
P(wx) = 13 + 14 = 27
2.59
Dist [v] = 18
s(0)
Iteration 5
s(0)
v(18)
z(8)
8
13
10
z(8)
11
w(13)
6
17
14
w(13)
7
16
y(15)
y(15)
x(20)
5
Figure 2.192
x(20)
EXAMPLE 2.83
Apply Dijkstra algorithm to find the shortest path
from s to t in the graph given below (Figure 2.193).
dist [s] = 0
dist [z] = 8
dist [w] =13
dist [y] =15
dist [v] =18
dist [x] = 20,
a
b
9
28
18
s
6
t
10
14
36
15
c
d
7
Figure 2.193
which are the required shortest paths from s to any
other point. The Dijkstra tree is shown in dark lines
in Figure 2.192.
Solution. Let x be the labelled endpoint of the edge
e, then P values are given by P(e) = dist [x] + w(e),
where dist [x] denotes the distance of x from s and
w(e) is the edge weight of e.
For each vertex v, dist [v] appears in the bracket. Iteration tree at the end of each iteration is shown by
dark lines:
Iteration 1
a
9
Iteration 4
28
18
s(0)
a(18)
b
6
14
36
15
c(15)
7
d
b(27)
18
t
10
9
s(0)
28
6
14
t
10
36
15
c(15)
7
d(22)
2.60

Engineering Mathematics-II
dist [s] = 0
P(s a) = 18 (minimum)
dist [s] = 0
P(d t) = 22 + 36 = 58
dist [c] = 15
P(c a) = 15 + 6 = 21
dist [c] = 15
P(b t) = 27 + 28 = 55
P(c b) = 15 + 14 = 29
(minimum)
P(c d) = 15 + 7 = 22
dist [a] = 18
P(c t) = ∞
dist [d] = 22
dist [b] = 27
Iteration 5
Iteration 2
a(18)
b
9
a(18)
18
s(0)
28
6
c(15)
dist [s] = 0
a(18)
9
b
28
6
14
t
10
t(55)
10
14
36
15
c(15)
Iteration 3
18
28
6
d
7
P(a b) = 18 + 9 = 27
P(c b) = 15 + 14 = 29
P(c d) = 15 + 7 = 22 (minimum)
P(c t) = ∞
P(a t) = ∞
dist [c] = 15
dist [a] = 18
s(0)
s(0)
36
15
b(27)
18
t
10
14
9
7
d(22)
dist [s] = 0
dist [c] = 15
dist [a] = 18
dist [s] = 0
dist [c] = 15
dist [a] = 18
dist [d] =22
dist [b] = 27
dist [t] =55
36
15
c(15)
7
d(22)
dist [s] = 0 P(c b) = 15 + 14 = 29
dist [c] = 15 P(a b) = 18 + 9 = 27
(minimum)
dist [a] = 18 P(d b) = 22 + 10 = 32
dist [d] = 22 P(d t) = 22 + 36 = 58
a(18)
Hence, the length of the shortest path (s, a, b, t)
from s to t is 55 and the Dijkstra’s tree is shown in
Figure 2.194.
b(27)
t(55)
s(0)
c(15)
d(22)
Figure 2.194
Graphs
EXAMPLE 2.84
Find a shortest path from s to t and its length for the
graph given below:
a
2
2
1
s
2
1
c
P(c) = dist [x] + w(e),
where dist [x] denotes the distance from s to x and
w(e) is the weight of the edge e.
For each vertex v, dist [v] appears in the bracket.
Iteration tree at the end of each iteration is shown in
dark lines.
t
1
d
Figure 2.195
Iteration 1
a
Iteration 2
b
3
a(2)
2
2
t
c(1)
dist [s] = 0
dist [c] = 1
1
Iteration 3
3
dist [s] =0
dist [c] = 1
dist [a] = 2
Iteration 4
a(2)
1
3
b
2
t
2
1
s(0)
2
1
1
d
P(a b) = 2 + 3 = 5
P(c d) = 1 + 1 = 2 (minimum)
P(d t) = ∞
P(b t) = ∞
P(a d) = 3
b
2
c(1)
1
c(1)
2
s(0)
t
2
1
d
P(c d) = 2
P(s a) = 2 (minimum)
P(s b) = ∞
P(s t) = ∞
P(d t) = ∞
P(b t) = ∞
a(2)
2
1
s(0)
2
1
b
3
2
1
s(0)
t(4)
2
1
c(1)
d(2)
1
d(2)
dist [s] = 0 P(d t) = 2 (minimum)
dist [s] =0
P(a b) = 5 (minimum)
dist [c] = 1
P(a b) = 5
dist [c] = 1
P(b t) = ∞
dist [a] = 2
P(b t) = ∞
dist [a] = 2
dist [d] = 2
2.61
Solution. Let x be the labelled endpoint of edge e,
then P values are given by
b
3

dist [d] = 2
dist [t] =4
2.62

Engineering Mathematics-II
dist [s] = 0
Iteration 5
dist [c] = 1
a(2)
dist [a] = 2
2
1
s(0)
dist [t] = 4
t(4)
2
1
c(1)
dist [b] = 5
b(5)
t(4)
s(0)
1
d(2)
Output: vk, vk − 1, …, v1, 0.
Else increase i by 1. Go to Step 3.
End Moore.
Thus, the Dijkstra tree is
c(1)
b(5)
2
dist [d] = 2
a(2)
3
d(2)
Figure 2.196
Remark 2.17. There could be several shortest
paths from s to t.
EXAMPLE 2.85
Use BFS algorithm to find shortest path from s to t in
the connected graph G given in Figure 2.197.
Thus the shortest path is scdt and its length is 4.
b(2)
2.16.2 Shortest Path if All Edges Have Length 1
If all edges in a connected graph G have length 1,
then a shortest path v1 → vk is the path that has the
smallest number of edges among all paths v1 → vk in
the given graph G.
Moore’s Breadth First Search Algorithm
This method of finding shortest path in a connected graph G from a vertex s to a vertex t is used
when all edges have length 1.
Input: Connected graph G = (V, E), in which one
vertex is denoted by s and one by t and each edge
(vi, vj) has length 1.
Initially all vertices are unlabelled.
Output: A shortest path s → t in G = (V, E).
1. Label s with 0.
2. Set vi = 0.
3. Find all unlabelled vertices adjacent to a vertex
labelled vi.
4. Label the vertices just found with vi + 1.
5. If vertex t is labelled, then “back tracking” gives
the shortest path. If k is label of t(i.e., t = vk), then
c(3)
a(1)
s(0)
t(3)
d(2)
e(2)
f (1)
Figure 2.197
Solution. Label s with 0 and then label the adjacent vertices with 1. Thus two vertices have been
labelled by 1. Now label the adjacent vertices of
all vertices labelled by 1 with label 2. Thus three
vertices have been labelled with 2. Label the vertices adjacent to these vertices (labelled by 2)
with 3. Thus two vertices have been labelled with
3. We have reached t. Now backtracking yields the
following three shortest paths:
(i) t(3), e(2), f (1), s(0), that is, s f e t,
(ii) t(3), b(2), a(1), s(0), that is s a b t,
(iii) t(3), e(2), a(1), s(0), that is, s a e t.
Graphs
Thus there are three possible shortest paths of
length 3.
EXAMPLE 2.86
Find a shortest path s → t in the graph given in Figure 2.198 (using Moore’s BFS algorithm).
t(5)
a(1)
h(4)
g(3)
s(0)
i(5)
c(3)
d(2)
Figure 2.198
Solution. Label s with 0 and then label the adjacent vertices with 1. Thus two vertices have been
marked as 1. Label the adjacent vertices of al vertices labelled as 1 with 2. Thus two vertices have
been marked as 2. Now label the adjacent vertices
of all vertices labelled as 2 with 3. Thus two vertices have been marked as 3. Label the adjacent vertices of all vertices labelled as 3 with 4. Thus three
vertices have been marked as 4. Now there is only
one vertex t to be labelled. Label it with 5. We have
reached at t.
Now backtracking gives the following shortest
paths:
or
or
or
sabght
sebcit
sabcit
Prim Algorithm
Prim algorithm builds a minimal spanning tree T by
expanding outward in connected links from some
vertex. In this algorithm, one edge and one vertex
are added at each step. The edge added is the one of
least weight that connects the vertices already in T
with those not in T.
Input: A connected weighted graph G with n
vertices.
Output: The set of edges E in a minimal spanning
tree.
1. Choose a vertex v1 of G. Let V = {v1} and E = { }.
2. Choose a nearest neighbour vi of V that is adjacent to vj, vj ∈ V and for which the edge (vi, vj)
does not form a cycle with members of E. Add vi
to V and add (vi, vj) to E.
3. Repeat Step 2 till number of edges in T is
n – 1. Then V contains all n vertices of G and E
contains the edges of a minimal spanning tree
for G.
Definition 2.114. A greedy algorithm is an algorithm
that optimizes the choice at each iteration without
regard to previous choices.
For example, Prim algorithm is a greedy
algorithm.
EXAMPLE 2.87
Find a minimal spanning tree for the graph shown
in Figure 2.199.
with length 5,
of length 5,
of length 5.
Thus, there are three possible shortest paths of
length 5.
2.17
2.63
f(4)
e(1)
thgbas
ticdes
ticbas
2.17.1

MINIMAL SPANNING TREE
Definition 2.113. Let G be a weighted graph. A spanning tree of G with minimum weight is called minimal spanning tree of G.
We discuss two algorithms to find a minimal
spanning tree for a weighted graph G.
a
4
2
b
11
7
8
c
d
6
1
f
9
e
Figure 2.199
Solution. We shall use Prim algorithm to find the
required minimal spanning tree. We note that
2.64

Engineering Mathematics-II
number of vertices in this connected weighted graph
is 6. Therefore the tree will have five edges.
We start with any vertex, say c. The nearest
neighbour of c is f and (c, f ) does not form a cycle.
Therefore (c, f ) is the first edge selected.
Now we consider the set of vertices V = {c, f }.
The vertex a is nearest neighbour to V = {c, f } and
the edge (c, a) does not form a cycle with the member of set of edges selected so far. Thus,
E = {(c, f ), (c, a)}
and
V = {c, f, a}.
The vertex b is now nearest neighbour to V = {c, f,
a} and the edge (a, b) do not form a cycle with the
member of E = {(c, f ), (c, a)}. Thus,
E = {(c, f ), (c, a), (a, b)} and V = {c, f, a, b}.
Now the edge (b, c) cannot be selected because it
forms a cycle with the members of E. We note that d
is the nearest point to V = {c, f, a, b} and (c, d) is the
edge which does not form a cycle with members of
E = {(c, f ), (c, a), (a, b)}. Thus we get
E = {(c, f ), (c, a), (a, b)}, (c, d)},
V = {c, f, a, b, d}.
The nearest vertex to V is now e and (d, e) is the corresponding edge. Thus,
E = {(c, f ), (c, a), (a, b), (c, d), (d, e)},
V = {c, f, a, b, d, e}
Since number of edges in the Prim tree is 5, the
process is complete. The minimal spanning tree is
shown in Figure 2.200:
a
4
2
b
8
c
d
6
1
f
e
Figure 2.200
The length of the tree is 1 + 4 + 2 + 8 + 6 = 21
EXAMPLE 2.88
Build the lowest-cost road system that will connect
the cities in the graph given in Figure 2.201.
B
2
H
3
3
A
D
6
2
2
5
C
5
4
6
3
E
4
F
G
Figure 2.201
Solution. We shall use Prim algorithm to build this
system. Here the number of vertices (Cities) is 8.
So, the minimal spanning tree would have seven
edges.
We start from any vertex, say A and have at the
initial stage:
V = {A}, E = { }.
The nearest neighbour of V = {A} is C and the corresponding edge is (A, C). So we have
E = {(A, C)},
V = {A, C}.
Now B is the nearest neighbour of V = {A, C} and
edge (C, B) does not form cycle with the edges in
E = {(A, C)}. Here the edge (A, B) can also be chosen. Thus, choosing arbitrarily one out of (A, B) and
(C, B), we have
E = {(A, C), (C, B)},
V = {A, C, B}.
Now, E and F are both nearest neighbour to V and
(C, E) or (C, F) do not form cycle with the members
of E = {(A, C), (C, B)}. So we may choose any of E
or F. Let us choose F and the corresponding edge
(C, F). Then,
E = {(A, C), (C, B), (C, F)},
V = {A, C, B, F}.
Now E and G are nearest neighbours of V = {A, C,
B, F} and both (F, E) and (F, G) do not form cycle
with the members of E = {(A, C), (C, B), (C, F)}.
Therefore we may pick up any of these vertices. We
pick arbitrarily E.
Solution. Pick up the vertex a. Then
Then,
E = {(A, C), (C, B), (C, F), (F, E)},
V = {A, C, B, F, E}.
E={ }
Now D is the nearest neighbour of V = {A, C, B, F,
E} and (E, D) is the corresponding edge to be added
to the tree. So, we have
E = {(A, C), (C, B), (C, F), (F, E), (E, D)},
V = {A, C, B, F, E, D}.
Now H is the nearest neighbour and (D, H) is the
corresponding edge and we have
E = {(A, C), (C, B), (C, F), (F, E), (E, D), (D, H)},
V = {A, C, B, F, E, D, H}.
Now G is the nearest neighbour and (E, G) is the
corresponding edge. So, we have
E = {(A, C), (C, B), (C, F), (F, E), (E, D),
(D, H), (E, G)},
V = {A, C, B, F, E, D, H, G}
Since the number of edges in the Prim tree is 7, the
process is complete. The minimal spanning tree is
shown in Figure 2.202:
B
D
H
2
3
A
2.65

Graphs
2
2
C
5
4
The nearest neighbour of V is b, d or e and the corresponding edges are (a, b) or (a, d) or (a, e). We
choose arbitrarily (a, b) and have
E = {(a, b)},
E = {(a, b), (a, d)},
E = {(a, b), (a, d), (d, e)},
Now c is the nearest neighbour of V = {a, b, d, e} and
the corresponding edges are (e, c), (d, c). Thus we
have, choosing (e, c),
E = {(a, b), (a, d), (d, e), (e, c)}, V = {a, b, d, e, c}
Total weight = 3 + 3 + 1 + 2 = 9.
(If we choose (d, c), then total weight is
3 + 3 + 1 + 2 = 9).
The minimal tree is shown in Figure 2.204.
4
a
3
1
2
c
2
d
Figure 2.203
c
or
a
1
3
EXAMPLE 2.89
Using Prim algorithm, find the minimal spanning
tree of the following graph (Figure 2.203):
e
3
2
2 + 3 + 5 + 4 + 2 + 2 + 3 = 21.
3
b
3
a
The total length of the roads is
3
V = {a, b, d, e}.
b
Figure 2.202
b
V = {a, b, d}.
Now e is the nearest neighbour of {a, b, d} and (d, e)
does not form cycle with {(a, b), (a, d)}.
Hence,
G
F
V = {a, b}.
Now d is the nearest neighbour of V = {a, b} and the
corresponding edge (a, d) does not form cycle with
(a, b). Thus we get
3
E
and V = {a}.
c
1
3
d
2
d
Figure 2.204
2.17.2
Kruskal’s Algorithm
In Kruskal’s algorithm, the edges of a weighted
graph are examined one by one in order of increasing
weight. At each stage, an edge with least weight out
of edge-set remaining at that stage is added provided
this additional edge does not create a circuit with
the members of existing edge set at that stage. After
n – 1 edges have been added, these edges together
with the n vertices of the connected weighted graph
form a minimal tree.
2.66

Engineering Mathematics-II
Algorithm
Input: A connected weighted graph G with n vertices and the set E = {e1, e2, …, ek} of weighted edges
of G.
The edge (v3, v5) has minimum weight, so we pick it
up. We have thus
Output: The set of edges in a minimal spanning
tree T for G.
The edges (v1, v4) and (v1, v5) have minimum weight
in the remaining edge set. We pick (v1, v4) say, as it
does not form a cycle with E*. Thus
Step 1. Initialize T to have all vertices of G and no
edges.
Step 2. Choose an edge e1 in E of least weight. Let
E* = {e1},
E = E − {e1}.
Step 3. Select an edge ei in E of least weight that
does not form circuit with members of E*.
Replace
E* by E* ∪ {ei}
E with E − {ei}.
and
Step 4. Repeat Step 3 until number of edges in E*
is equal to n – 1.
EXAMPLE 2.90
Use Kruskal’s algorithm to determine a minimal
spanning tree for the connected weighted graph G
shown in Figure 2.205.
E* = {(v2, v4), (v3, v5)},
E = E − {(v2, v4), (v3, v5)}.
E* = {(v2, v4), (v3, v5), (v1, v4),
E = E − {(v2, v4), (v3, v5), (v1, v4)}.
Now the edge (v1, v5) has minimum weight in
E − {(v1, v4), (v3, v5), (v1, v4)} and it does not form a
cycle with E*. So, we have
E* = {(v2, v4), (v3, v5), (v1, v4), (v1, v5)}
and
E = E − {(v2, v4), (v3, v5), (v1, v4), (v1, v5)}.
Thus all the four edges have been selected. The minimal tree has the edges.
(v2, v4), (v3, v5), (v1, v4), (v1, v5)
and is shown in the Figure 2.206.
v1
v1
4
3
5
3
3
v2
7
v3
6
2
v4
v3
2
2
2
5
3
v2
v5
v4
(Minimal spanning tree)
v5
Figure 2.206
Figure 2.205
Solution. The given weighted graph has five vertices.
The minimal spanning tree would, therefore, have
four edges.
Let
EXAMPLE 2.91
Using Kruskal’s algorithm, find the minimal spanning tree for the graph given in the Figure 2.207.
E = {(v1, v2), (v1, v4), (v1, v5), (v2, v3), (v1, v3),
(v2, v4), (v4, v5), (v5, v3), (v3, v4)}.
The edges (v2, v4) and (v3, v5) have minimum weight.
We choose arbitrarily one of these, say (v2, v4).
Thus
E* = {(v2, v4)},
E = E − {(v2, v4)}.
a
2
2
6
e
4
3
c
1
3
3
b
d
5
f
Figure 2.207

Graphs
Solution. The given graph has six vertices, therefore
the minimal spanning tree would have 6 – 1 = 5
edges
Let
Remark 2.18. In the above example, if we had chosen (e, f ) in place of (c, f ) in the last step, then the
minimal spanning tree would have been as shown in
the Figure 2.209.
E = {(a, b), (a, c), (b, d ), (c, f ), (d, f ), (c, d ), (a, e),
(e, f ), (e, c)}.
2
2
e
E* = {(c, d), (a, c),
E = E − {(c, d), (a, c)}.
Now choose the edge (a, e), since it has also minimum weight in E − {(c, d),(a, c)} and it does not
form cycle with E*. So
3
d
1
c
3
f
(Minimal spanning tree)
E = E − {(c, d)}.
Now choose the edge (a, c) because it has minimum
weight. So,
b
a
Choose the edge (c, d) first because it has minimum
weight. Therefore,
E* = {(c, d)},
2.67
Figure 2.209
EXAMPLE 2.92
Use Kruskal’s algorithm to find a shortest spanning
tree for the graph G given in the Figure 2.210.
a1
E* = {(c, d), (a, c), (a, e)}, E = E − {(c, d),
(a, c), (a, e)}.
3
a3
a2
4
1
6
2
5
2
a4
3
a5
Now choose (b, d) since it has minimum weight and
it does not form circuit with E*. Thus
Figure 2.210
E* = {(c, d ), (a, c), (a, e), (b, d )},
E = E − {(c, d ), (a, c), (a, e), (b, d )}.
Solution. The given graph has five vertices. So the
required spanning tree would have 5 – 1 = 4 edges.
Let
Now choose (c, f) or (e, f) because they have equal
and minimal weight. Let us choose (c, f ). Then
E = {(a1, a2), (a1, a3), (a3, a4), (a2, a4), (a2, a5), (a4,
a5), (a2, a3), (a1, a4)}.
E* = {(c, d ), (a, c), (a, e), (b, d ), (c, f)},
E = E − {(c, d ), (a, c), (a, e), (b, d ), (c, f)}.
Choose the edge (a2, a3) since it has minimum
weight. Then
E* = {(a2, a3)},
E = E − {a2, a3}
= {(a1, a2), (a1, a3), (a3, a4), (a2, a4),
(a2, a5), (a4, a5), (a1, a4)}.
We have thus obtained all the five required edges. The
minimal spanning tree is shown in the Figure 2.208.
b
a
2
2
c
1
3
e
f
(Minimal spanning tree)
Figure 2.208
3
d
Now choose the edge (a2, a4) or (a3, a4) having minimum weight in E. Let us take (a2, a4), say. Then
E* = {(a2, a3), (a2, a4)}
and
E = E − {(a2, a3), (a2, a4)}
= {(a1, a2), (a1, a3), (a3, a4),
(a2, a5), (a4, a5), (a1, a4)}.
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
Engineering Mathematics-II
We cannot choose (a3, a4) now because it will form
circuit with already chosen edges (a2, a3), (a2, a4).
Therefore we can choose (a1, a3) or (a4, a5). Suppose
we choose (a1, a3). Then
E* = {(a2, a3), (a2, a4), (a1, a3)},
E = E − {(a2, a3), (a2, a4), (a1, a3)}
= {(a1, a2), (a3, a4), (a2, a5), (a4, a5), (a1, a4)}.
Now choose (a4, a5) as it has minimal weight in E.
Then
E* = {(a2, a3), (a2, a4), (a1, a3), (a4, a5)}
and
E = E − {(a2, a3), (a2, a4), (a1, a3), (a4, a5)}
= {(a1, a2), (a3, a4), (a2, a5), (a1, a4)}.
d(a, e) = 3, d(a, c) = 2, d(b, e) = 2 and diam (G) = 3.
Definition 2.116. A vertex v in a connected graph G
is called a cut point if G – v is disconnected, where
G − v is the graph obtained from G by deleting v and
all edges containing v.
For example, in the above graph, d is a cut point.
Definition 2.117. An edge e of a connected graph G
is called a bridge (or cut edge) if G – e is disconnected, where G – e is the graph obtained by deleting
the edge e.
For example, consider the graph G shown in the
Figure 2.213.
Thus we have obtained four edges and therefore the
process is complete.
The Kruskal minimal spanning tree is shown in
the Figure 2.211.
a1
a
e4
c
e5
2
a3
a4
e
G
a5
3
Figure 2.213
Figure 2.211
Its length is 3 + 1 + 2 + 3 = 9.
2.18
b
e3
d
a2
1
3
e1
e2
We observe that G – e3 is disconnected. Hence the
edge e3 is a bridge.
CUT SETS
Let G be a connected graph. We know that the distance between two vertices v1 and v2, denoted by
d(v1, v2), is the length of the shortest path.
Definition 2.115. The diameter of a connected graph
G, denoted by diam (G), is the maximum distance
between any two vertices in G.
For example, in graph G shown in the Figure
2.212, we have
a
Definition 2.118. A minimal set C of edges in a connected graph G is said to be a cut set (or minimal
edge-cut) if the subgraph G – C has more connected
components than G has.
For example, in the above graph, if we delete
the edge (b, d) = e3, the resulting subgraph G – e3
has two connected components shown in the
Figure 2.214.
b
e1
d
c
a
e2
d
b
e4
e
G
Figure 2.212
and
c
Figure 2.214
e5
e
Graphs
EXAMPLE 2.93
Find a cut set for the graph given below in the
Figure 2.215.
v2
e2
v2
e8
e4
v5
e3
e2
v4
v5 e6
v1
v3
e8
v1
e5
e7
2.69
e5, e6, e7. The two connected components are shown
in the Figure 2.216.
So, in this example, the cut set consists of single
edge (b, d) = e3, which is called cut edgeor bridge.
e1

v3
Figure 2.216
e3
But, if we remove any proper subset of {e1, e4,
e5, e6, e7}, then there is no increase in connected
components of G. Hence {e1, e4, e5, e6, e7} is a cut set.
v4
Figure 2.215
Solution. The given graph is connected. It is sufficient to reduce the graph into two connected components. To do so we have to remove the edges e1, e4,
EXAMPLE 2.94
Find a cut set for the graph given in the Figure 2.217.
f
b
e1
e2
e6
e7
e5
a
e3
g
e
d
e4
e8
e9
h
c
G
Figure 2.217
The given graph G is connected. Clearly G – e5 has two connected components give in the Figure 2.218.
f
b
e2
e1
e6
and
a
d
e3
e7
g
e
e4
e8
e9
h
c
Figure 2.218
2.70

Engineering Mathematics-II
Hence a cut set for this graph is {e5}. Some other
cut sets for G are {e1, e3}, {e3, e2}, {e6, e9}, {e6, e8}.
EXAMPLE 2.95
Find a cut set for the graph represented by the
Figure 2.219.
Solution. We note that {e4, e5} is a cut set (minimal
edge cut) for the given graph because removal of e4
and e5 gives rise to two components shown in the
Figure 2.222.
e9
g
f
b
e1
a
e7
e2
e6
e7
e8
d
e9
e5
e3
e4
g
e10
,
d
e3
e9
g
e4
e11
h
c
Figure 2.220
Hence the set {e7, e10} is a cut set for the given
graph G.
EXAMPLE 2.96
Find a cut set (minimal edge cut) for the following
graph (Figure 2.221).
g
e10
e5
j
e1
b
i
e12
e11
e
e4
a
h
e9
e8
e6
c
e3
d
Figure 2.222
Similarly, {e9} forms a cut set (bridge) for this
graph.
Theorem 2.30. Let G be a connected graph with n
e8
e2
e6
e5
f
e2
f
b
e7
j
and
b
Solution. The given graph is a connected graph. We
note that removal of the edges e7 and e10 creates two
connected components of G shown in Figure 2.220.
a
e12
e1
h
Figure 2.219
e1
e11
e
a
e11
c
i
e8
e6
f
e10
h
e2
c
Figure 2.221
e3
d
vertices. Then G is a tree if and only if every edge of
G is a bridge (cut edge).
(This theorem asserts that every edge in a tree
is a bridge).
Proof: Let G be a tree. Then it is connected and has n –
1 edges (proved already). Let e be an arbitrary edge
of G. Since G – e has n – 2 edges, and also we know
that a graph G with n vertices has at least n – C(G)
edges, it follows that n – 2 ≥ n – C(G – e). Thus, G – e
has at least two components. Thus removal of the
edge e created more components than in the graph
G. Hence e is a cut edge. This proves that every edge
in a tree is a bridge.
Conversely, suppose that G is connected and
every edge of G is a bridge. We have to show that
G is a tree. To prove it, we have only to show that
G is circuit-free. Suppose on the contrary that there
exists a cycle between two points x and y in G (Figure 2.223). Then any edge on this cycle is not a cut
edge which contradicts the fact that every edge of G
is a cut edge. Hence G has no cycle. Thus G is connected and acyclic and so is a tree.
Graphs
y
x
Figure 2.223
2.18.1
Relation Between Spanning Trees, Circuits
and Cut Sets
A spanning tree contains a unique path between any
two vertices in the graph. Therefore, addition of a
chord to the spanning tree yields a subgraph that
contains exactly one circuit. For example, consider
the graph G shown below in the Figure 2.224.
v1

2.71
Then the set of e – v + 1 circuits obtained by adding
e – v + 1 chords to a spanning tree of G is called the
fundamental system of circuits relative to the
spanning tree.
A circuit in the fundamental system is called a
fundamental circuit.
For example, {v1, v2, v3, v1} is the fundamental
circuit corresponding to the chord (v1, v2).
On the other hand, since each branch of a tree
is cut edge, removal of any branch from a spanning tree breaks the spanning tree into two trees.
For example, if we remove (v1, v3) from the above
figured spanning tree, the resulting components are
shown in the Figure 2.226.
v1
v2
v3
v4
v5
v2
v3
v4
v5
G
Figure 2.224
Figure 2.226
For this graph, the Figure 2.225 is a spanning tree.
Thus, to every branch in a spanning tree, there
is a corresponding cut set. But, in a spanning tree,
there are v – 1 branches. Therefore, there are v – 1
cut sets corresponding to v – 1 branches.
v1
v2
v4
v3
v5
Figure 2.225 (Spanning tree)
The chords of this tree are (v1, v2) and (v2, v4). If we
add (v1, v2) to this spanning tree, we get a circuit
v1 v2 v3 v1. Similarly addition of (v2, v4) gives one
more circuit v2 v3 v5 v4 v2. If there are v vertices and
e edges in a graph, then there are e – v + 1 chords in
a spanning tree. Therefore, if we add all the chords
to the spanning tree, there will be e – v + 1 circuits in
the graph.
Definition 2.119. Let v be the number of vertices and e be the number of edges in a graph G.
Definition 2.120. The set of v – 1 cut sets corresponding to v – 1 branches in a spanning tree of a graph
with v vertices is called the fundamental system of
cut sets relative to the spanning tree.
A cut set in the fundamental system of cut sets is
called a fundamental cut set.
For example, the fundamental cut sets in the
spanning tree (figured above) are
{(v1, v2), (v1, v3)}, {(v1, v3), (v2, v3), (v3, v4)},
{(v3, v5), (v4, v5)}, {v2, v4), (v4, v5)}.
Theorem 2.31. A circuit and the complement of
any spanning tree must have at least one edge in
common.
2.72

Engineering Mathematics-II
Proof: We recall that the set of all chords of a tree
is called the complement of the tree. Suppose on
the contrary that a circuit has no common edge
with the complement of a spanning tree. This
means the circuit is wholly contained in the spanning tree. This contradicts the fact that a tree is
acyclic (circuit- free). Hence a circuit has at least
one edge in common with complement of a spanning tree.
Theorem 2.32. A cut set and any spanning tree must
have at least one edge in common.
Proof: Suppose on the contrary that there is a cut
set which does not have a common edge with a
spanning tree. Then removal of cut set has no effect
on the tree, that is, the cut set will not separate the
graph into two components. But this contradicts the
definition of a cut set. Hence the result.
Theorem 2.33. Every circuit has an even number of
edges in common with every cut set.
Proof: We know that a cut set divides the vertices of
the graph into two subsets each being set of vertices in one of the two components. Therefore a path
connecting two vertices in one subset must traverse
the edges in the cut set an even number of times.
Since a circuit is a path from some vertex to itself,
it has an even number of edges in common with
every cut set.
2.19
TREE SEARCHING
Let T be a binary tree of height h ≥ 1 and root v.
Since h ≥ 1, v has at least one child : vL and/or vR.
Now vL and vR are the roots of the left and right subtrees of v called TL and TR, respectively.
v Level 0
vL
vR
Level 1
Level 2
Level 3
Level 4
Left subtree TL
(Dotted circle)
Right subtree TR
(Dotted circle)
Figure 2.227
Definition 2.121. Performing appropriate tasks at a
vertex is called visiting the vertex.
Definition 2.122. The process of visiting each
vertex of a tree in some specified order is called
searching the tree or walking or traversing the
tree.
We now discuss methods of searching a tree.
Graphs
1. Pre-order Search Method
Input: The root v of a binary tree.
Output: Vertices of a binary tree using preorder
traversal
1. Visit v
2. If vL (left child of v) exists, then apply the algorithm to (T(vL), vL)
3. If vR (right child of v) exists, then apply this algorithm to (T(vR), vR)
End of algorithm pre-order.
In other words, pre-order search of a tree consists of the following steps:
2.73

If we apply pre-order algorithm to the given tree T, we
visit the root and print A. Using pre-order to left subtree, we visit the root B and print B and then search
TL printing C and E and then D. Up to this point, the
search has yielded the string A B C E D. Thus the
search of TL is complete. We now go to subtree TR.
Using the same procedure, we get the string F G H I.
Thus the complete search of the tree T is
A B C E D F G H I.
EXAMPLE 2.98
Find binary tree representation of the expression
(a − b) × (c + (d ÷ e))
Step 1. Visit the root.
Step 2. Search the left subtree if it exists.
Step 3. Search the right subtree if it exists.
and represent the expression in string form using
pre-order traversal.
EXAMPLE 2.97
Find the order of the vertices of the following tree T
processed using pre-order traversal.
Solution. In the given expression, × is the central
operator and therefore shall be the root of the binary
tree. Then the operator − acts as vL and the operator
+ acts as vR. Thus the tree representation of the given
expression is as shown in the Figure 2.230.
A
×
F
B
C
I
E
+
−
G
D
H
a
b
÷
c
e
T
d
Figure 2.228
Solution. The root of the given tree is A. It has two
children B and F. The left subtree and right subtree
are shown in Figure 2.229.
C
D
This form of the expression is called prefix form or
polish form of the expression
G
,
I
H
E
Left subtree TL
Right subtree TR
Figure 2.229
The result of the pre-order traversal to this binary
tree is the string
× − a b + c ÷ d e.
F
B
Figure 2.230
(a − b) × (c + (d ÷ e)).
In a polish form, the variables a, b, c,…are called
operands and −, + , × , ÷ are called operators. We
observe that, in polish form, the operands follow
the operator.
2.74

Engineering Mathematics-II
EXAMPLE 2.99
Represent the expression (A + B) * (C − D) as a binary
tree and write the prefix form of the expression.
–
C
+
Solution. Here * is the central operator. Further + and
− operators are vL and vR. Hence the binary tree for
the given experssion is as shown in the Figure 2.231.
B
A
Figure 2.232
EXAMPLE 2.101
Evaluate the polish form
*
× − 64 + 5 ÷ 22
−
+
Solution. We have the following steps in this regard:
A
B
C
D
Figure 2.231
Using pre-order traversal, the prefix expression
for it is
* + A B − C D.
2.19.1 Procedure to Evaluate an Expression Given
in Polish Form
To find the value of a polish form, we proceed as
follows:
Move from left to right until we find a string
of the form K x y, where K is operator and x, y are
operands.
Evaluate x K y and substitute the answer for the
string K x y. Continue this procedure until only one
number remains.
EXAMPLE 2.100
Find parenthesized form of the polish expression
− + A B C.
Solution. The parenthesized form of the given polish
expression is derived as follows:
1.
2.
3.
4.
5.
6.
7.
8.
× (6 − 4) + 5 ÷ 22
× 2 + 5 ÷ 22
× 2 + 5 (2 ÷ 2)
× 2 + 51
× 2 (5 + 1)
× 26
2×6
12, which is the required value of the expression.
2. Post-order Search Method
Algorithm
Step 1. Search the left subtree if it exists
Step 2. Search the right subtree if it exists
Step 3. Visit the root
End of algorithm
EXAMPLE 2.102
Represent the expression
(A + B) * (C − D)
as a binary tree and write the result of post-order
search for that tree.
Solution. The binary tree expression (as shown earlier) of the given algebraic expression is given in
Figure 2.233.
*
− (A + B) C,
(A + B) − C.
The corresponding binary tree is shown in the
Figure 2.232.
–
+
A
B
C
Figure 2.233
D
Graphs
The result of post-order search of this tree is
A B+C D−*
This form of the expression is called postfix form
of the expression or reverse polish form of the
expression.
In postfix form, the operator follows its
operands.
EXAMPLE 2.103
Find the parenthesized form of the postfix form
A B C * * C D E+/−.
Solution. We have
1.
2.
3.
4.
A B C * * C D E+/−
A (B * C) * C (D + E) / −
(A * (B * C )) (C / (D + E )) −
(A * (B * C )) − (C / (D + E )).
The corresponding binary tree is shown in the
Figure 2.234.
–
*
A
C
B
TRANSPORT NETWORKS
An important application of weighted directed graph
is to model transport networks like oil pipeline,
water supply pipeline, communication network,
electric power distribution system and highway
system. The purpose of a network is to implement
the flow of oil, water, electricity, messages, traffic,
etc and we desire that the flow through the network
should be largest possible value. Such a flow will be
called maximum flow.
Definition 2.123. A transport network or simply a
network is a simple weighted directed graph with
the following properties:
(i) There is a designated vertex (node), called the
source, that has no incoming edge.
(ii) There is a designated vertex, called the sink,
that has no outgoing edge.
(iii) There is a non-negative weight cij on the
directed edge (i, j), called the capacity of the
edge (i, j).
Definition 2.124. A flow in a network is a function that
assigns to each (i, j) of the network a non-negative
number fij such that
+
C
2.75
Since the graph is simple, it follows that if the edge
(i, j) is in the network, then ( j, i) is not there.
/
*
2.20

D
Figure 2.234
E
(i) 0 ≤ f ij ≤ cij , where cij is the capacity of the edge
(i, j)
(ii) For each vertex j, which is neither the source
nor the sink,
EXAMPLE 2.104
Evaluate the postfix form
∑f
i
21 − 342 ÷ + × .
Solution. We have
21 − 342 ÷ + × = (2 − 1) 342 ÷ + ×
= 13 (4 ÷ 2) + ×
= 132 + ×
= 1 (3 + 2) ×
= 15 ×
= 1 × 5 = 5.
ij
= ∑ f jk .
k
The non-negative number fij is called the flow in
the edge (i, j). For any vertex (distinct from source
and sink) j, the sum ∑ f ij is called the flow into j and
the sum
∑f
i
jk
is called the flow out of j. Thus the
k
condition (ii) implies that material cannot accumulate, be created, dissipate or lost at any vertex other
than the source or the sink. Hence the equation
2.76
∑f
i

ij
Engineering Mathematics-II
= ∑ f jk is called conservation of flow. As a
value of the flow and is denoted by value (F). A flow
F in a network is represented by labelling each edge
(i, j) with the pair (cij, fij).
For example consider the oil pipeline network
shown in Figure 2.235.
k
consequence of conservation, it follows that the sum
of flows leaving the source must be equal to the sum
of the flows entering the sink. This sum is called the
A
B
(4, 3)
(5, 5)
(5, 3)
(2, 1)
O (Dock)
(2, 1)
S (Refinery)
(5, 1)
(3, 3)
C
D
(4, 2)
Figure 2.235
Flow out of the vertex B = 5.
Therefore, conservation of flow is satisfied. Further,
value (F) = 3 + 3 = 5 + 1 = 6.
In this network, the crude oil unloaded at the
dock O (source) is being pumped to the refinery S
(sink). The capacities of the flow are
cOA = 5, c AB = 4, cBS = 5, cOC = 3,
EXAMPLE 2.105
Consider the water pipeline network shown in
Figure 2.236 for two towns A and B in which water
is supplied from four borewells w1, w2, w3, w4 and
a, b, c, d represent intermediate pumping stations.
Form an equivalent network with unique source
and unique sink and determine the value of the
flow.
cCB = 2, cCD = 4, cDB = 2, cDS = 5.
The flows in the edges are
f OA = 3,
f AB = 3,
f BS = 5,
f OC = 3,
fCB = 1,
fCD = 2,
f DB = 1, C DS = 1.
We note that
Flow into the vertex B = 3 + 1 + 1 = 5
w1
(4, 3)
(4, 4)
a
A
(2, 2)
(3, 2)
(2, 1)
w2
b
(5, 3)
w3
w4
(4, 2)
(5, 3)
c
(3, 2)
(4, 2)
d
Figure 2.236
B

Graphs
Solution. To obtain the required equivalent network
with designated source and designated sink, we tie
together the given sources w1, w2, w3 and w4 into a
w1
super source O and the towns A and B into a super
sink S as shown in Figure 2.237.
(4, 4)
a
(4, 3)
(2, 2)
(∞, 3)
(2, 1)
A
(3, 2)
(∞, 6)
b
(∞, 2)
w2
O
(Source)
2.77
S(sin k)
(∞, 3)
(5, 3)
(5, 3)
w3
(4, 2)
(∞, 4)
c
(∞, 2)
(3, 2)
w4
B
(4, 2)
d
Figure 2.237
The value of the flow is
Value (F) = 3 + 2 + 3 + 2 = 6 + 4 = 10.
out of five edges are carrying maximum capacity.
Therefore, the flow function in the second case is
better than the first one.
Definition 2.125. A maximal flow in a network N is a
flow with maximum value.
In general, there will be several flows having
the same maximum value. For example, consider the
network shown in Figure 2.238.
a
(7, 7)
(7, 5)
O
S
(2, 2)
a
(5, 3)
(5, 5)
7
7
b
O
Figure 2.239(a)
S
2
a
5
5
(7, 7)
(7, 7)
b
and
O
Consider the following two flows for this network:
The value of the flow in Figure 2.239(a) is 10 and
three of the five edges in this network attain their
maximum capacity. But for the same network, the
value of the flow in Figure 2.239(b) is 12 and four
S
(2, 0)
Figure 2.238
(5, 5)
(5, 5)
b
Figure 2.239(b)
2.78

Engineering Mathematics-II
EXAMPLE 2.106
Find the maximum possible increase in the following network (Figure 2.240).
A
O
(4, 3)
(2, 1)
D
B
Figure 2.240
Figure 2.243
Solution. We cannot increase the flow in the path
OABS because the capacity and the flow of the edge
AB are equal. But if increase flow in the path OCDS
by 1, we get the network shown in Figure 2.241.
A
(3, 3)
Solution. The capacity of each edge is shown in Figure 2.243. Let the initial flow be 0 in each edge. Thus
we have the network as shown in Figure 2.244(a).
A
B
(4, 3)
(4, 2)
(2, 1)
O
5
2
(3, 2)
(3, 2)
(2,0)
(4,0)
S
(2, 0)
O
(3, 3)
D
Further, if we increase the flow in the path OCBS by
1, we get the network shown in Figure 2.242.
(3, 3)
B
(2, 1)
B
Figure 2.244(a)
Increase the flow by 2 in the path OAS to get the
network shown in the Figure 2.244(b).
(4, 4)
(4, 2)
O
A
S
(2, 1)
(3, 3)
(5, 5)
C
(5,0)
(2,0)
Figure 2.241
A
S
(3,0)
(3, 3)
(5, 4)
C
S
3
S
(2, 0)
(5, 3)
C
2
4
B
(3, 3)
(4, 2)
O
A
(3, 3)
(2,2)
(4,2)
D
Figure 2.242
No further increase in flow is possible since the
capacities of all the edges leading to the sink have
been exhausted. The value of the flow is 7.
EXAMPLE 2.107
Find a maximal flow in the network shown in
Figure 2.243.
O
S
(3,0)
(5,0)
(2,0)
B
Figure 2.244(b)

Graphs
Now increase the flow by 2 in the path OBS to get
the network shown in Figure 2.244(c).
A
E
3
2.79
B
4
3
4
O
A
S
1
5
2
(2,2)
(4,2)
C
2
D
Figure 2.245
O
S
(3,0)
Solution. The initial labelling yields the network
shown in Figure 2.246(a).
(5,2)
(2,2)
B
A
Figure 2.244(c)
(4,0)
Now increase the flow by 2 in the path OABS to get
the network shown in Figure 2.244(d).
(3,0)
B
(4,0)
E
(3,0)
O
S
(1,0)
(5,0)
(2,0)
A
C
(2,0)
D
Figure 2.246(a)
(2,2)
(4,4)
O
S
(3,2)
Increase the flow by 2 in the path OCDS by to get
the network shown in Figure 2.246(b).
(5,4)
(3,2)
B
Figure 2.244(d)
Any further increase in the flow is not possible.
Thus maximal flow in the network has been reached.
Thus, value (F) = 2 + 4 = 6.
EXAMPLE 2.108
Find a maximal flow in the network shown in
Figure 2.245.
A
(4,0)
(3,0)
B
(4,0)
E
O
(3,0)
S
(1,0)
(5,2)
(2,2)
C
(2,2)
D
Figure 2.246(b)
Now increase the flow by 3 in the path OEABS to
get the network shown in Figure 2.246(c).
2.80

Engineering Mathematics-II
A
(4,3)
For example, the capacity of the cut in Figure
2.247 is
B
(3,3)
(4,3)
E
(3,3)
O
EXAMPLE 2.109
Find the capacity of the cut, shown by dashed line,
in the network shown in Figure 2.248.
(5,2)
(2,2)
C
C ( P , P ) = COC + C AC + C AB = 5 + 2 + 3 = 10.
S
(1,0)
D
(2,2)
Figure 2.246(c)
a
(2,2)
No further increase in flow in possible since the
capacities in the edges (A, B) and (C, D) has been
exhausted. Hence value (F) = 5.
A
B
(3,3)
(4,3)
(4,2)
(2,0)
O
S
(2,1)
(3,2)
(5,3)
C
(3,2)
D
Figure 2.247
Definition 2.127. The capacity of a cut K = ( P , P )
is the sum of the capacities of all edges in K and is
denoted by C(K) or C ( P , P ) .
(4,2)
(2,0)
(1,0)
(1,1)
O
c
(2,0)
(3,1)
(3,2)
Definition 2.126. Let N be a network. A cut in N is a
partition ( P , P ) of the set of vertices in N such that
the source O ∈ P and the sink S ∈ P .
Thus, a cut ( P , P ) in a network N is a set K of
edges ( v, w ), v ⑀ P , w ⑀P such that every path from
source to sink contains at least one edge from K. In
fact, a cut does cut a directed graph into two pieces,
one containing the source and the other containing
the sink. Nothing can flow from source to sink if
edges of a cut are removed. We indicate a cut by
drawing a dashed line to partition the vertex set. As
an illustration, consider the network of EXAMPLE
2.106. The dashed line divides the vertex set into the
sets P = {O, A} and P = {C,B,D,S } .
b
(3,2)
d
(3,1)
f
(1,0)
(3,2)
S
(2,2)
e
Figure 2.248
Solution. The cut ( P , P ) is given by
P = {o, a, c} , P = {d , e, f , b, S } .
Therefore the capacity of the given cut is
C ( P , P ) = cOd + ccf + Cab = 3 + 1 + 3 = 7.
Definition 2.128. A minimal cut is a cut having minimum capacity.
Theorem 2.34. (The Max Flow Min Cut Theorem) A
maximum flow F in a network has value equal to the
capacity of a minimum cut of the network.
Proof: Let F be any flow and K be any cut in a network. Since all parts of F must pass through the
edges of the cut K and since C(K) is the maximum
amount that can pass through the edges of K, it follows that value ( F ) ≤ C ( K ) . The equality will hold
if the flow F uses the full capacity of all edges in the
cut K. Thus equality holds if the flow is maximum.
Further, K must be a minimum capacity cut since
every cut must have capacity at least equal to value
(F). This completes the proof of the theorem.
EXAMPLE 2.110
In the network shown in Figure 2.249, find a
maximum flow, give its value and prove that it is
Graphs

2.81
maximum by appealing to the max flow min cut
theorem.
Now increase the flow by 1 in the path OCS to get
the network shown in Figure 2.250(c).
A
A
(2,1)
1
2
(1,1)
1
3
O
2
(1,0)
4
B
S
O
1
1
(3,0)
(2,1)
B
(4,0)
S
(1,0) (1,1)
C
C
Figure 2.249
Figure 2.250(c)
Solution. The initial labelling yields the following
network shown in Figure 2.250(a):
Further, increase in flow by 1 in the path OABS to
get the network of Figure 2.250(d).
A
A
O
(2,2)
(1,0)
(2,0)
(1,1)
(1,1)
(1,0)
(3,0)
(2,0)
B
(4,0)
S
O
(3,0)
B
(4,1)
(1,1)
(1,0)
(1,0)
S
(2,1)
(1,0)
C
C
Figure 2.250(a)
Figure 2.250(d)
Increase the flow by 1 in the path OAS to get the
network shown in Figure 2.250(b).
Now increase the flow by 1 in the path OCBS and
get the network shown in Figure 2.250(e).
A
A
(2,1)
(2,2)
(1,1)
(1,0)
O
(3,0)
B
(4,0)
(1,1)
S
O
(3,0)
B
(4,2)
(1,1)
(2,0)
(1,0) (1,0)
(2,2)
C
Figure 2.250(b)
(1,1)
C
Figure 2.250(e)
S
(1,1)
2.82

Engineering Mathematics-II
Lastly, we increase the flow by 2 in the path OBS
and get the network with maximum flow as shown
in Figure 2.250(f). In fact the capacities in the edges
leading to the sink have been completely exhausted.
A
(2,2)
O
B
(4,4)
(1,1)
(iv) K 4 = {OC , OB, AS , AB}
with cap ( K 4 ) = COC + COB + C AS + C AB
= 2 + 3 + 1 + 1 = 7.
We note that capacity of the minimum cut in 6.
Hence, the maximum flow is 6.
(1,1)
(1,1)
(3,2)
(iii) K 3 = {CS , BS , AS }
with cap( K 3 ) = CCS + C BS + C AS = 1 + 4 + 1 = 6,
EXERCISES
S
(1,1)
(2,2)
C
Figure 2.250(f)
We note that value (F = 2 + 2 + 2 = 1 + 4 + 1 = 6).
To check whether it is maximum flow, we appeal to
max flow min cut theorem. We consider the following cuts:
1. Is there a non-empty simple graph with twice as
many edges as vertices?
Ans. Yes. In fact if n = number of vertices
2n = number of edges, then using the formula
for number of edges in a complete graph:
n( n − 1)
n( n − 1)
which
, we have 2n =
2
2
yields n = 5. But in a complete graph K5 the
number of edges is 10. Thus the complete
graph K5, which is simple by definition, has
five vertices and ten edges.
2. Is the graph given below a bipartite graph?
v1
A
v3
2
1
3
O
v2
B
1
4
S
v5
v4
Figure 2.251
2
1
1
C
Figure 2.250(g)
(i) K1 = {OC , OB, OA}
with cap
( K1 ) = COC + COB + COA = 2 + 3 + 2 = 7,
(ii) K 2 = {OC , BC , BS , AS }
with cap
( K 2 ) = COC + C BC + C BS + C AS
= 2 + 1 + 4 + 1 = 8,
Ans. Yes, it is a bipartite graph, where the
set of vertices V has been partitioned into
two subsets, V1 = {v3, v4}, V2 = {v1, v2, v5}.
But it is not a complete bipartite graph.
3. Find a formula for the number of edges in complete bipartite graph Kmn.
Ans. The m + n vertices having been divided
into two subsets V1 and V2 containing m vertices and n vertices, respectively. Every verticex
in V1 is joined to all the n vertices in V2. Thus
the task of joining each vertex of V1 to each
vertex in V2 can be performed in mn ways.
Hence the number of edges in Kmn is mn.
4. A graph G has vertices of degrees 1, 4, 3, 7, 3 and
2. Find the number of edges in the graph.

Graphs
Ans. Total degree of G is 1 + 4 + 3 + 7 + 3 +
2 = 20 which should be twice the number of
edges. Hence number of edges in G is 10.
5. How many edges are there in an n-cube?
Ans. Degree of each vertex in an n-cube in n.
The number of vertices in an n cube in 2n.
Therefore total degree of n-cube in n2n. Hence
the number of edges is an n-cube is
n ⋅ 2n
= n ⋅ 2n −1.
2
6. When does the complete graph Kn possess an
Euler circuit?
Ans. Complete graph is a connected graph.
So it will have an Euler cycle if degree of
each vertex is even.
7. Does the graph given below have an Euler
circuit?
v2
v1
v5
v1
v2
Figure 2.252
Ans. Yes. The graph is connected and degree
of each vertex is even. Hence the graph possesses an Euler circuit. For example, v1 v2 v5
v4 v5 v2 v3 v4 v1 is an Euler cycle.
8. When does the complete bipartite graph Kmn
contain Euler cycle?
Ans. When both m and n are even. In such a
case the degree of each vertex will be even.
9. In seven bridges problem, was it possible for
a citizen of Konigsberg to make a tour of the
city and cross each bridge exactly twice? Give
reasons.
Ans. Yes, because in such a case degree of
each vertex of the graph of seven bridges problem shall be even.
10. Show that the graph given below does not contain a Hamiltonian cycle.
v3
v5
v4
v7
v6
v9
v8
v10
Figure 2.253
11. Give example of a graph that has Euler circuit
but not Hamiltonian circuit.
d
Ans.
a
c
e
b
12. Give example of a graph that has an Hamiltonian
circuit but not an Euler circuit.
v1
Ans.
v3
v4
2.83
v2
v3
13. Give example of a graph that has both an Euler
circuit and an Hamiltonian circuit.
Ans.
K3
14. Use Dijkstra’s shortest path algorithm to find the
shortest path from a to f in the graph given below:
e
3
1.7
a
2
5.4
b
3
2.3
3.3
d
2.2
f
1.5
c
Figure 2.254
Ans. 6.7
2.84

Engineering Mathematics-II
15. Find the adjacency matrix of the graph.
v1
18. Defining an isomorphism. Show that the graphs
shown below are isomorphic.
x
b
w
c
a
v3
v2
y
z
d
G1
G2
x
v5
v4
Figure 2.255
Ans.
⎡0
⎢1
⎢
⎢1
⎢
⎢0
⎢⎣0
z
Figure 2.256
1 1 0 0⎤
0 0 1 1⎥
⎥
0 0 1 1⎥
⎥
1 1 0 1⎥
1 1 1 0 ⎥⎦
16. Find directed graph that have the following
adjacency matrix.
⎡0
⎢1
⎢2
⎢⎣1
Ans.
0
0
1
1
1
2
0
0
y
w
(Clearly, the graph G2 can be drawn as)
1⎤
0⎥
0⎥ .
0 ⎥⎦
Ans. Vertex set of G1 is V(G1) = {a, b, c, d}
vertex set of G2 is V(G2) = {x, y, z, w}
Define f : V(G1) → V(G2) by
f (a) = y, f (b) = w, f (c) = x, f (d) = z.
Then f is bijective.
19. Show that the graphs shown below are not
isomorphic. Give reasons.
a
y
d
e
x
z
b
c
t
n
m
G2
f
G1
v1
v2
Figure 2.257
Ans. Graph G2 is connected whereas G1 is
not connected. The connected components
of G1 are
v3
v4
17. A graph G has the adjacency matrix given
below. Verify whether it is connected.
⎡0
⎢0
A = ⎢0
⎢0
⎣⎢0
0
0
0
1
0
1
0
0
0
1
0
1
0
0
1
0⎤
0⎥
1⎥ .
1⎥
0 ⎦⎥
20. Show that the graphs shown below are not
isomorphic.
a
x
b
c
d
e
y
z
n
Ans. The graph G is connected because B = A +
A2 + A3 + A4 has non-zero entries off the
main diagonal.
f
G1
G2
Figure 2.258
m
Graphs
Ans. The graphs do not have same number
of vertices and so are not isomorphic.
21. Find the complement of the graph shown below.
a
d
e
b
c
f
Figure 2.259
Ans. The complement of the given graph is
a
b
c
d
e

2.85
Ans. We note that
(i) d(v1, v2) = 1, d(v1, v3) = 2, d(v1, v4) = 3
and so e(v1) = 3
(ii) d(v2, v1) = 1, d(v2, v3) = 1, d(v2, v4) = 2
and so e(v2) = 2
(iii) d(v3, v1) = 2, d(v3, v2) = 1, d(v3, v4) = 1
and so e(v3) = 2
(iv) d(v4, v1) = 3, d(v4, v2) = 2, d(v4, v3) = 1
and so e(v4) = 3
Hence rad (G) = min {3, 2, 2, 3} = 2. The
centre is {v2, v3}.
23. Show that any graph having n vertices, n = 1, 2,
3, 4 is planar.
24. Show that the colouring of map given below
requires at least three colours
A
B
C
D
E
F
G
f
Figure 2.261
22. Let G be a graph and V be its vertex set. Then
(i) Eccentricity of v ∈ V is defined as
e(v) = max {d(u, v): u ∈ V, u ≠ v}.
(ii) The radius of the graph is defined by
rad(G) = min {e(v): v ∈ V}.
(iii) The diameter of G is defined by
diam (G) = max {e(v): v ∈ V}.
= max [d(u, v): u ∈ V,
v ∈ V, u ≠ v}.
(iv) v is a central point if
e(v) = rad(G).
(v) The centre of G is the set of all central
points.
Find the centre of the graph given below:
v1
v2
v3
v4
Figure 2.260
Ans. Give one colour to A and D, give second colour to B, E and G, and third colour to
C and F. Thus all the adjoining boundaries
have different colours.
25. How many colours are required to paint the
graph given below?
v2
v7
v6
v1
v5
v3
v4
Figure 2.262
Ans. Three. We can paint v1, v2, v3 with
colours C1, C2, C3. Then paint v4 by C1, v5 by
C2, v6 by C3 and v7 by C1.
26. For which values of m and n is the complete
bipartite graph Kmn a tree?
2.86

Engineering Mathematics-II
Ans. Number of edges in Kmn = mn. Total number of vertices in Kmn = m + n. If kmn is a tree, then
number of edges in Kmn should be (m + n) − 1.
Hence, to be a tree kmn satisfies the condition
mn = (m + 1) − 1 or m + n = mn + 1,
which is satisfied if either m = 1 or n = 1.
27. Let G be a graph with n vertices and n − 2 or
fewer edges. Show that G is not connected.
Ans. Suppose G is connected. The graph
itself will be a tree or it can be made a tree
by eliminating some edges. In either case it
must have n − 1 edges. But in the questions,
number of edges is less than or equal to n − 2.
Hence G cannot be connected.
28. Represent the weighted graph given below in
the matrix form.
Ans.
−
/
⋅
a
+
c
b
d
e
Prefix form: − • ab −c + de
Postfix form: ab • de + c /−
30. Convert the following postfix form of a binary
tree into prefix form and parenthesized infix
form and usual infix form
ab + c − .
−
Ans.
v1
v6
1
4
4
3
a
v3
5
4
b
v2
2
v5
c
+
1
3
3
2
v4
Prefix form: − + abc
Parenthesized form: (a + b) − c
Usual infix form: a + b −c
31. Draw a tree for the algebraic expression
(3x + y) (4m + 5n)2
Figure 2.263
and find its prefix polish form.
Ans.
⎡∞ 1 4 1 3 3 ⎤
⎢ 1 ∞ 2 ∞ 4 ∞⎥
⎢
⎥
⎢ 4 2 ∞ 2 ∞ ∞⎥
⎢
⎥
⎢1 ∞ 2 ∞ 4 5⎥
⎢3 4 ∞ 4 ∞ 3⎥
⎢
⎥
⎢⎣ 3 ∞ ∞ 5 3 ∞ ⎥⎦
The diagonal is ∞ ∞ … ∞ and the matrix is
symmetric with respect to the diagonal.
29. Represent the expression given below by a
binary tree and obtain the prefix form and postfix form of the expression.
a·b − (c/(d + e)).
Ans. Using ↑ for exponentiation and * for
multiplication, the binary tree representing
the algebraic expression is
∗
+
↑
y
∗
2
+
3
x
4
∗
∗
m
5
n
Ans.
The prefix form is
* + * 3xy ↑ + * 4m * 5n
232* ↑ + 233 − 24 = * ↑ (2 + 3) 3 (2 − 4)
= * (2 + 3)3 (2 − 4)
= (2 + 3)3 * (2 − 4)
= 53 * (−2)
= −250
32. If ↑ denotes exponentiation, evaluate the polish form
* ↑ + 233 − 24.
33. Find all cut sets for the graph G given
below.
f
b
c
a
e
a
7
4
c
b
3
10
e
d
35. Find a minimal tree for the graph of Exercise
34 by deleting one by one those costliest edges
whose deletion does not disconnect the graph.
Ans. First remove {a, d}, then {d, c}, {a, e},
{b, e}. We cannot remove {c, e} now because
the graph will be disconnected by doing so.
But we can delete {b, d}. Now delete any of
{a, c} or {b, c}. We will get two minimal trees.
36. Find all spanning trees of the graph shown below:
a
g
d
2.87

Graphs
i
b
h
d
c
Figure 2.264
Figure 2.266
Ans. {c, e}; {{a, b}, {a, d}}; {{a, d}, {b,
c}}; {{e, f }, {e, g}} {h, g}}; {g, i}; {{e,
h},{h, g}}. Try for other cut edges.
34. Find a minimal spanning tree for the graph
shown below:
Ans. Each spanning tree must have n − 1 =
4 − 1 = 3 edges. There are eight such trees.
37. Find the value of the maximum flow in the network of EXAMPLE 2.109 and verify the value
obtained by max flow min cut theorem.
Ans. Value (F) for max flow = 7, minimum
cut (k) is shown in Figure 8.248.
38. Find the value of the maximum flow in the network shown in Figure 2.267 and verify your
answer using max flow min cut theorem.
a
A
7
4
4
b
c
8
d
10
3
12
4
2
15
Figure 2.265
e
S
4
5
C
12
B
3
O
20
2
2
D
Figure 2.267
Ans. Max flow = 6
This page is intentionally left blank.
3
3.1
Improper Integrals
IMPROPER INTEGRAL
While evaluating Riemann integrals, the integrand
was assumed to be bounded and the range (interval)
of integration was assumed to be finite. The aim of
this chapter is to consider the integrals in which the
above conditions are not satisfied. This leads us to a
concept called the improper integral.
Definition 3.1. If the integrand f is unbounded or the
limits a or b or both are infinite, then the integral
Rb
f ðxÞdx is called an improper integral, generalized
a
integral or an infinite integral.
For example, the integrals
R1
(i) dx
x3 ;
0
(ii)
(iii)
R2
dx
ð1xÞð2xÞ
1
R1 dx
1þx2
1
and
are improper integrals. We note that in integral (i),
the integrand becomes infinite (unbounded) at the
lower limit 0. In such a case, we say that x ¼ 0 is
a point of infinite discontinuity (singular point) of
the integrand f. In integral (ii), both lower and upper
limits are singular points of f, whereas in integral
(iii), the range of integration is infinite.
An integral that is not improper is called proper
integral. In view of Definition 3.1, we shall
consider the following two cases:
(a) When the integral is unbounded.
(b) When the range of integration is infinite.
3.2
CONVERGENCE OF IMPROPER INTEGRAL
WITH UNBOUNDED INTEGRAND
In what follows, we assume that the number of
singularities of the integrand in any interval of
integration is finite.
Definition 3.2. If the lower limit is the only singularity
of the integrand f, then the improper integral
Rb
f ðxÞdx is said to converge at a if
a
Zb
f ðxÞdx
lim
m!0þ
aþm
exists and is finite.
EXAMPLE 3.1
Z2
dx
.
Examine the convergence of
x2
0
Solution. We note that the lower limit 0 is the only
singularity of the integrand x12 in the interval
½0; 2. Then
Z2
Z2
dx
dx
¼ lim
;
0 < m < 2;
x2 m!0þ
x2
m
0
¼ lim
m!0þ
1 1
¼ 1:
m 2
Since the limit is not finite, the given improper integral is divergent.
Definition 3.3. If the upper limit is the only singularity
of the integrand f, then the improper integral
Rb
f ðxÞdx is said to converge at b if
a
Zbm
f ðxÞdx
lim
m!0þ
a
exists and is finite.
EXAMPLE 3.2
Z1
Examine the convergence of
0
dx
ð1 xÞ2
.
3.2
n
Engineering Mathematics-II
Solution. Since the upper limit 1 is the only singularity of the integrand, we have
Z1
0
Z1m
dx
ð1 xÞ
2
¼ lim
m!0þ
0
dx
; 0<m<1
ð1 xÞ2
1 1m
m!0þ 1 x 0
1
¼ lim 1 ¼ 1:
m!þ m
3.2.1
COMPARISON TESTS
(A) Let f and be two functions such that
f ðxÞ 0; ðxÞ 0 and f ðxÞ ðxÞ for all
x 2 ½a; b. Then
Rb
Rb
(i) f ðxÞdx converges if ðxÞdx converges.
a
¼ lim
(ii)
Definition 3.4. If both upper and lower limits are the
only singularities of the integrand, and c is any
point within the interval of integration ½a; b, then
Rb
the improper integral f ðxÞdx converges if both
improper integrals
a
Rb
f ðxÞdx and
a
a
ðxÞdx diverges if
a
Hence, the given improper integral is divergent.
Rc
Rb
f ðxÞdx converge.
c
Moreover,
Rb
f ðxÞdx diverges.
a
Proof. Assume that f and are both bounded and
integrable in the interval ½a þ m ; b ; 0 < m ðb aÞ and a is the only singularity. Then, by the
given hypothesis,
Zb
Zb
f ðxÞdx aþm
ðxÞdx;
m e ð0 ; b aÞ: ð1Þ
aþm
Rb
To prove (i), let ðxÞdx converge so that there
a
exists a number M such that
Zb
Zc
Zb
f ðxÞdx ¼
a
Zb
f ðxÞdx þ
a
ðxÞdx < M for all m 2 ð0; b aÞ:
f ðxÞdx:
aþm
c
Hence,
EXAMPLE 3.3
R2
Examine the convergence of
dx
xð2xÞ :
0
Solution. Both 0 and 2 are singularities of the integrand. Therefore,
Z2
dx
¼
xð2 xÞ
0
Z1
dx
þ
xð2 xÞ
0
1
¼
2
Z1
0
1
1
1
dx þ
x 2x
2
Z2
dx
xð2 xÞ
Zb
f ðxÞdx < M for all m 2 ð0; b aÞ
aþm
Rb
and so
f ðxÞdx converges at a.
Rb
To prove (ii), we note that if f ðxÞdx does not
a
Rb
converge at a, then
f ðxÞdx is not bounded
a
aþm
1
Z2
1
1
dx
x 2x
1
h
h
1
x i1 1
x i2m
¼ lim log
þ lim log
2 m!0þ
2 x m 2 m!0þ
2x 1
1
m
¼ 0 lim log
2 m!0þ
2m
1
2m
þ 0 ¼ 1:
þ lim log
2 m!0þ
m
Hence, the given improper integral diverges.
above. Consequently, (1) implies that
Rb
ðxÞdx is
aþm
also not bounded above. Hence,
Rb
ðxÞdx does not
a
converge.
(B) If f and g are two positive functions such that
f ðxÞ
¼ l and l is neither 0 nor 1, then the
lim gðxÞ
x!aþ
Rb
Rb
two integrals f ðxÞdx and gðxÞdx converge or
a
diverge together.
a
Improper Integrals
f ðxÞ
Proof. Under the given hypothesis, gðxÞ
is positive for
f ðxÞ
all x and so l cannot be negative. Since lim gðxÞ
¼ l,
x!aþ
there exists a neighbourhood ½a; c ; a < c < b
such that for all x 2 ½a; c,
f ðxÞ
<e
l
gðxÞ
(i) Convergence of
Rb
f ðxÞdx:
a
(ii) Divergence of
Rb
Rb
n
3.3
gðxÞdx ) convergence of
a
Rb
f ðxÞdx ) divergence of
a
gðxÞdx.
a
or
Remark 3.1.
f ðxÞ
<lþe
le<
gðxÞ
f ðxÞ
x!aþ gðxÞ
(i) If lim
or
Rb
ðl eÞgðxÞ < f ðxÞ < ðl þ eÞgðxÞ
ð2Þ
f ðxÞ < ðl þ eÞgðxÞ
ð3Þ
Now, due to relation (2),
Zb
f ðxÞdx
Convergence of
a
Zc
)
f ðxÞ
x!aþ gðxÞ
a
Zc
) ðl eÞ
gðxÞdx converges (using Test (A))
a
Zc
gðxÞdx converges
a
Zb
)
¼ 1 and
Rb
gðxÞdx diverge, then
a
f ðxÞdx also diverges.
a
EXAMPLE 3.4
Examine the convergence of
Zb
dx
ðComparison IntegralÞ
ðx aÞn
a
f ðxÞdx converges
)
f ðxÞdx also converges.
(ii) If lim
Rb
and
gðxÞ converges, then
a
a
or
ðl eÞgðxÞ < f ðxÞ
Rb
¼ 0 and
gðxÞdx converges at a.
Solution. For n 6¼ 1, we have
Zb
Zb
dx
dx
n ¼ lim
m!0þ
ðx aÞ
ðx aÞn
a
aþm
"
#b
1
¼ lim
m!0þ ð1 nÞðx aÞn1
aþm
"
#
1
1
1
lim
¼
1 n m!aþ ðb aÞn1 mn1
a
¼
Using (3), it may similarly be shown that
Zb
gðxÞdx does not converge
a
Zb
)
1
1n ðb
1
If n ¼ 1, then we have
Zb
Zb
dx
dx
¼ lim
x a m!0þ
xa
a
f ðxÞdx does not converge.
a
Also from (3), it follows that
aÞ1n
if n < 1
if n > 1.
aþm
¼ lim ½logðx aÞbaþm
m!0þ
¼ lim ½logðb aÞ log mÞ ¼ 1:
m!0þ
3.4
n
Engineering Mathematics-II
Hence, the given integral converges if n < 1 and
diverges if n 1.
EXAMPLE 3.5
R1
Examine the convergence of
0
dx
.
x1=3 ð1þx2 Þ
1
1
Solution. Let f ðxÞ ¼ x1=3 ð1þx
2 Þ and gðxÞ ¼ x1=3 . Then
f ðxÞ
1
¼ lim
¼ 1 (finite):
lim
x!0 gðxÞ
x!0 1 þ x2
1
1
R
R
Therefore
f ðxÞdx and
gðxÞdx converge or
0
R1 0
R1
diverge together. But f ðxÞdx ¼ xdx
1=3 is con0
0
vergent since 13 < 1 (see Example 3.4). Hence,
R1
R1
dx
f ðxÞdx ¼ x1=3 ð1þx
2 Þ also converges.
0
0
EXAMPLE 3.6
R1
Examine the convergence of
0
0
Z1=2
dx
x1=2 ð1
xÞ
1=2
¼
0
Hence,
1=2
xÞ
Z1
dx
x1=2 ð1
xÞ1=2
ð4Þ
For the first integral on the right side of (4), let
1
1
f ðxÞ ¼
and gðxÞ ¼ 1=2 :
1=2
x
x1=2 ð1 xÞ
f ðxÞ
¼ 1 (finite)
x!0 gðxÞ
1=2
1=2
R
R
and so
f ðxÞdx and
gðxÞdx behave alike. But
0
0
1=2
R
gðxÞdx converges since 12 < 1.
the integral
lim
0
1=2
R
0
dx
also converges. Since both
x1=2 ð1xÞ1=2
1=2
integrals on the right-hand side of (4) converge, it
follows that the given integral is also convergent.
EXAMPLE 3.7
Examine the convergence of
Z1
bðm; nÞ ¼ xm1 ð1 xÞn1 dx ðBeta FunctionÞ
Solution.
Case II. If m and n are less than 1, then the integrand
has 0 and 1 as the points of infinite discontinuity. In
that case, we have
Z1=2
Z1
x
m1
dx
x1=2 ð1xÞ1=2
xm1 ð1 xÞn1 dx
dx ¼
0
Z1
xm1 ð1 xÞn1 dx:
þ
ð5Þ
1=2
For the first integral on the right side of (5), we take
f ðxÞ ¼ xm1 ð1 xÞn1 ¼
¼
also converges.
For the second integral on the right side of (4), let
1
1
and ðxÞ
:
f ðxÞ ¼
1=2
1=2
x ð1 xÞ
ð1 xÞ1=2
ð1 xÞ
n1
0
Then
Hence, the integral
R1
ð1xÞ
Case I. If m and n both are greater than or equal to 1,
the integrand is finite for all values of x from 0 to 1.
Hence, in that case, the given integral is convergent.
dx
1=2
1=2
0
x1=2 ð1
þ
f ðxÞ
¼ 1 (finite)
x!1 gðxÞ
1
R1
R
f ðxÞdx and
ðxÞdx behave alike.
Therefore,
1=2
1=2
R1
dx
1
But the integral
1=2 converges since 2 < 1.
lim
(For a detailed study on Beta Function, refer
Chapter 4.)
dx
.
x1=2 ð1xÞ1=2
Solution. The integrand is unbounded at both limits
0 and 1. Therefore, we may write
Z1
Then
1
x1m
ð1 xÞn1
and gðxÞ
x1m
:
Then
f ðxÞ
¼ lim ð1 xÞn1 ¼ 1 (finite):
x!0 gðxÞ
x!0
lim
n
Improper Integrals
1=2
R
f ðxÞdx and
gðxÞdx converge or
0
1=2
1=2
R 1
R 0
gðxÞdx ¼
diverge together. But
x1m dx con-
Therefore,
1=2
R
0
EXAMPLE 3.8
R1
Evaluate
Solution. By definition, we have
Z1
0
positive.
To test the convergence at 1 of the second integral on the right side of (5), we take
xm1
and
f ðxÞ ¼ xm1 ð1 xÞn1 ¼
ð1 xÞ1n
1
:
ðxÞ ¼
ð1 xÞ1n
Then
f ðxÞ
¼ lim xm1 ¼ 1 (finite)
lim
x!1 ðxÞ
x!1
R1
R1
Therefore,
f ðxÞdx and
ðxÞdx converge or
1=2
1=2
R1
ðxÞdx ¼
diverge together. But the integral
dx
1þx2 .
0
0
verges if and only if 1 m < 1 if and only if m > 0.
1=2
R
f ðxÞdx converges at 0 if and only if m is
Hence,
dx
¼ lim
1 þ x2 b!1
0
1=2
Zb
dx
¼ lim ½tan1 xb0
1 þ x2 b!1
0
:
2
¼ lim tan1 b ¼
b!1
Definition 3.6. Let f be bounded and integrable in
Rb
½t; b, where t b. If lim f ðxÞdx is finite,
t!1 t
Rb
then the improper integral
1
converge at 1.
Thus, by definition,
Zb
f ðxÞdx;
t! 1
1
ð1xÞ1n
1
dx converges if and only if 1 n < 1 if
and only if n is positive. Thus, it follows from (5)
R1
that xm1 ð1 xÞn1 dx exists (converges) for all
0
positive values of m and n.
f ðxÞdx is said to
Zb
f ðxÞdx ¼ lim
1=2
R1
3.5
t
provided that the limit exists and is finite.
EXAMPLE 3.9
R0
Evaluate
1
dx
1þx2.
Solution. We have
3.3
Z0
CONVERGENCE OF IMPROPER INTEGRAL
WITH INFINITE LIMITS
1
dx
¼ lim
1 þ x2 t! 1
Z0
dx
1 þ x2
t
¼ lim ½tan1 x0t
Definition 3.5. Let f be bounded and integrable in
½a; b, where b is any number greater than or equal
Rb
to a. If lim f ðxÞdx is finite, then the improper
t!1
¼
:
2
b!1 a
integral
R1
f ðxÞdx is said to converge at 1.
a
Thus, by definition,
Zb
Z1
f ðxÞdx ¼ lim
f ðxÞdx;
b!1
a
0
provided that the limit exists and is finite.
Definition 3.7. Let f be bounded and integrable in
½a; b, where b a. If t is any number and both
R1
Rt
f ðxÞdx and f ðxÞdx converge at 1 and 1,
1
t
respectively, then the improper integral
R1
1
f ðxÞdx
is said to converge at both limits 1 and 1.
3.6
n
Engineering Mathematics-II
Thus, by definition, for any number t
Zt
Z1
Z1
f ðxÞdx ¼
f ðxÞdx þ
f ðxÞdx;
1
1
1
provided that the integrals on the right exist.
3.3.1
COMPARISON TESTS
(A) If 0 f ðxÞ gðxÞ for all x in ½a; x, then
R1
R1
(i) f ðxÞdx converges if gðxÞdx converges.
a
R1
(ii)
a
gðxÞdx diverges if
a
EXAMPLE 3.10
R1
Evaluate
1
f ðxÞ
¼ l,
(B) If f and g are positive in ½a; x and lim gðxÞ
x!1
where l is a non-zero finite number, then the
R1
R1
integrals f ðxÞdx and gðxÞdx converge or
dx
1þx2.
a
dx
¼
1 þ x2
1
¼
EXAMPLE 3.11
Show that
Z1
dx
;
xn
Z0
1
f ðxÞdx diverges.
a
Solution. We have
Z1
R1
dx
þ
1 þ x2
Z1
a
diverge together. Further, if lim
dx
1 þ x2
0
þ ¼ by Examples 3.8 and 3.9.
2 2
f ðxÞ
x!1 gðxÞ
R1
gðxÞdx converges, then
a
and if lim
f ðxÞ
x!1 gðxÞ
then
R1
¼ 1 and
R1
¼ 0 and
f ðxÞdx converges
a
R1
gðxÞdx diverges,
a
f ðxÞdx diverges.
a
a > 0 ðComparison IntegralÞ
a
converges if and only if n > 1:
(The test (B) is called the Quotient Test).
(C) p-test: If lim xp f ðxÞ ¼ l, then
x!1
R1
(i) f ðxÞdx converges if p > 1 and l is finite.
(ii)
a
R1
f ðxÞdx diverges if p 1 and l is non-zero.
a
Solution. When n 6¼ 1, we have
Zb
dx
1
1
1
¼
n1 :
n
n1
x
1n b
a
a
Therefore,
1
Zb
lim
b!1
2
dx
1
1
1
¼ lim
n1
n
n1
b!1 1 n b
x
a
a
¼
1
ðn1Þan1
1;
;
n>1
n<1
When n ¼ 1, we have
Zb
EXAMPLE 3.12
Examine the convergence of
Z1
2
ex dx ðEulerPoisson IntegralÞ:
dx
b
¼ ½log xba ¼ log ! 1
x
a
Solution. Since ex > x2 for all real x, we have
R1 1
2
ex < x12 . Since
x2 dx is convergent, by the
1
comparison test, the given Euler–Poisson integral is
also convergent.
EXAMPLE 3.13
R1
Examine the convergence of
as b ! 1:
1
Solution. Let
a
Thus, the given integral converges if n > 1 and
diverges if n 1.
dx
.
x1=3 ð1þxÞ1=2
f ðxÞ
1
x1=3 ð1 þ xÞ
1=2
and gðxÞ
1
x1=3 x1=2
¼
1
x5=6
:
Improper Integrals
R1
R1
¼ 1. Hence, f ðxÞdx and gðxÞdx
1
1
R1 dx
behave identically. But x5=6 diverges since 56 < 1.
1
R1
dx
Hence,
is
also
divergent.
1=2
1=3
f ðxÞ
x!1 gðxÞ
Then lim
x
1
ð1þxÞ
R1
f ðxÞdx is
Definition 3.8. The improper integral
a
R1
called absolutely convergent if jf ðxÞjdx converges.
If
then
R1
R1
a
f ðxÞdx converges but
a
R1
jf ðxÞjdx diverges,
a
f ðxÞdx is called conditionally convergent.
a
f ðxÞ
¼ lim ex
x!0 gðxÞ
x!0
R1 1
¼ x1p converges
0 R1
lim
and
R1
gðxÞdx
0
p > 0. Therefore
cos x
x2 þ1 dx
is absolutely convergent.
cos x 1 and
Solution. Since 2
x þa1 1 þ x2
Z1
Z
dx
dx
¼
lim
¼ lim ½tan1 xa0 ¼ ;
1 þ x2 a!1
1 þ x2 a!1
2
0
0
1
R dx
the integral 1þx
2 is convergent. Therefore, by the
0
R1 cos x comparison test, the integral
1þx2 dx is con0
0
For the integral
0
(For a detailed study on Gamma Function, refer
Chapter 4.)
Solution. Let
R1
f ðxÞdx, we examine the
convergence at 1. So, let gðxÞ ¼ x12 so that
f ðxÞ
xpþ1
¼ lim x ¼ 0 for all p:
x!0 gðxÞ
e
R1 dx
R1
But x2 converges. Therefore, f ðxÞdx converges
lim
1
for all p.
Therefore, from (6), it follows that the Gamma
Function converges for all positive values of p.
EXERCISES
1. Examine the convergence of
R1
dx
(i) x1=3 ð1þx
2Þ
0
Ans. Convergent
=2
R
f ðxÞ ¼ x e :
If p < 1, then 0 is a point of infinite discontinuity
of f . Then
Z1
Z1
Z1
f ðxÞdx ¼ f ðxÞdx þ
f ðxÞdx:
ð6Þ
For the integral
0
1
f ðxÞdx, we examine the con-
1
vergence at 0. So, let gðxÞ ¼ xp1 ¼ x1p
. Then
R1
(iii)
emx xn dx
0
Ans. Converges for n > 1
2. Examine the convergence of
R1 x
(i) log
x2 dx
1
(ii)
0
log sin x dx
Ans. Convergent
p1 x
R1
if 1 p < 1 or if
0
EXAMPLE 3.15
Examine the convergence of
Z1
ex xp1 dx ðGamma FunctionÞ
0
¼1
1
(ii)
vergent. Hence, the result.
3.7
f ðxÞdx converges if p > 0.
0
1
EXAMPLE 3.14
R1
Show that
n
R1
0
(iii)
R1
0
(iv)
R1
0
Ans. Convergent
2
sin x
x2 dx
Ans. Convergent
p1
x
1þx dx
Ans. Converges for 0 < p < 1
sin xm
xp
dx
Ans. Converges if 1 m < p < 1 þ m
R1 x
3. Show that sin
xn is absolutely convergent.
1
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4
Beta and Gamma Functions
Za
Za
The beta and gamma functions, also called Euler’s
Integrals, are the improper integrals, which are
extremely useful in the evaluation of integrals.
f ðxÞdx ¼
0
f ða xÞdx;
0
Z1
4.1
The integral
R1
xm1 ð1 xÞn1 dx, which converges
0
¼ bðn; mÞ; m > 0; n > 0:
0
for m > 0 and n > 0 is called the beta function and is
denoted by b (m, n). Thus,
Z1
xm1 ð1 xÞn1 dx; m > 0; n > 0:
bðm; nÞ ¼
Thus,
bðm; nÞ ¼ bðn; mÞ; m > 0; n > 0:
2. We have
Z1
Beta function is also known as Eulerian Integral of
First Kind.
As an illustration, consider the integral
R1 1
x2 ð1 xÞ4 dx. We can write this integral as
0
Z1
3
x 21 ð1 xÞ51 dx;
0
0
:2 sin h cos h dh
which is a beta function, denoted by b 32 ; 5 . But, on the
R1 1
other hand, the integral x2 ð1 xÞ3 dx is not a beta
Z
2
¼2
sin2m2 h cos2n2 h sin h cos h dh
0
0
function as n 1 ¼ 3 implies n ¼ 2 (negative).
Z2
¼2
PROPERTIES OF BETA FUNCTION
1. We have
sin2m1 h cos2n1 h dh:
0
Thus,
Z1
Z1
xm1 ð1 xÞ
n1
dx
xm1 ð1 xÞn1 dx
bðm; nÞ ¼
0
0
Z1
ð1 xÞm1 ð1 ð1 xÞÞn1 dx; since
¼
0
ð1Þ
Putting x ¼ sin2 h, we get dx ¼ 2sin h cos h dh and
therefore, (1) becomes
Z2
bðm; nÞ ¼ ðsin2 hÞm1 ð1 sin2 hÞn1
0
bðm; nÞ ¼
xm1 ð1 xÞn1 dx:
bðm; nÞ ¼
0
4.2
xn1 ð1 xÞm1 dx
¼
BETA FUNCTION
Z2
¼2
sin2m1 h cos2n1 h dh:
0
4.2
n
Engineering Mathematics-II
3. Let m and n be positive integers. By definition,
Z1
xm1 ð1 xÞn1 dx; m > 0; n > 0:
bðm; nÞ ¼
t
1
4. Put x ¼ 1þt
so that dx ¼ ð1þtÞ
2 dt. Then the
expression for beta function reduces to
m1 n1
Z1 t
t
1
bðm; nÞ ¼
1
:
dt
1þt
1þt
ð1 þ tÞ2
0
Integration by parts yields
bðm; nÞ ¼ xm1
ð1 xÞn
nð1Þ
Z1
1
0
ðm nÞxm2
0
m1
¼
n
Hence, if m and n are positive integers, then
ðm 1Þ!ðn 1Þ!
bðm; nÞ ¼
:
ðm þ n 1Þ!
0
Z1
ð1 xÞn
dx
nð1Þ
¼
0
Z1
x
m2
Z1
n
ð1 xÞ dx
¼
0
0
m1
¼
bðm 1; n þ 1Þ:
n
tm1
dt ¼
ð1 þ tÞmþn
Z1
bðm; nÞ ¼
m2
bðm 2; n þ 2Þ
nþ1
1
bð1; m þ n 1Þ:
bð2; m þ n 2Þ ¼
mþn2
bðm 1; n þ 1Þ ¼
0
ðm 1Þðm 2Þ:::::ð2Þð1Þ
nðn þ 1Þðn þ 2Þ:::::ðm þ n 2Þ
Z1
x11 ð1 xÞmþn2 dx
0
Z1
0
xm1
dx þ
ð1 þ xÞmþn
0
"
ðm 1Þ! ðn 1Þ! ð1 xÞmþn1
¼
ðm þ n 1Þð1Þ
ðm þ n 2Þ!
ðm 1Þ! ðn 1Þ!
1
¼
:
ðm þ n 2Þ! m þ n 1
Z1
dx
¼
#1
0
Z1
1
xm1
dx
ð1 þ xÞmþn
ð2Þ
¼ I1 þ I2 ; say:
In I2, put x ¼ 1t so that dx ¼ t12 dt and so,
Z1
Z0 1m1 1
tmþn
1
t
2 dt ¼
: dt
I2 ¼ m1
1 mþn
t
t ðt þ1Þmþn t2
1þ t
1
ð1 xÞ
ðm 1Þ! ðn 1Þ!
:
¼
ðm þ n 1Þ!
0
0
¼
mþn2
xm1
dx:
ð1 þ xÞmþn
0
Z1
ðm 1Þ! ½1:2:3::::ðn 1Þ
1:2::::ðn 1Þnðn þ 1Þðn þ 2Þ:::ðm þ n 2Þ
ðm 1Þ! ðn 1Þ!
¼
ðm þ n 2Þ!
0
xm1
dx:
ð1 þ xÞmþn
Since b(m, n) ¼ b(n, m), we have
Z1
Z1
xm1
xn1
dx
¼
dx:
bðm; nÞ ¼
mþn
ð1 þ xÞ
ð1 þ xÞmþn
bð1;m þ n 1Þ
Z1
2. From the property (4), we have
Z1
xm1
dx
bðm; nÞ ¼
ð1 þ xÞmþn
Multiplying the preceding equations, we get
¼
1
1
:
:
dt
ð1 þ tÞm1 ð1 þ tÞn1 ð1 þ tÞ2
Hence,
Similarly,
bðm;nÞ ¼
tm1
0
n1
t
dt ¼
ð1þtÞmþn
Z1
0
0
n1
x
dx:
ð1þxÞmþn
Hence, (2) reduces to
Z1
Z1
xm1
xn1
dx
bðm; nÞ ¼
mþn dx þ
ð1 þ xÞ
ð1 þ xÞmþn
0
Z1
¼
0
0
xm1 þ xn1
dx; m > 0; n > 0:
ð1 þ xÞmþn
Beta and Gamma Functions
EXAMPLE 4.1
Show that 1
Z
0
0
0
Z1
ð1xÞm xn1 dx; since bðmþ1;nÞ ¼ bðn;mþ1Þ
¼
m1
x
dx:
ð1 þ xÞmþn
ð3Þ
0
¼ ð1xÞ
Since b(m, n) ¼ b(n, m), we have
Z1
xn1
dx:
bðm; nÞ ¼
ð1 þ xÞmþn
0
Adding (3) and (4), we get
Z1 m1
x
þ xn1
dx:
2bðm; nÞ ¼
ð1 þ xÞmþn
ð4Þ
¼
m
n
¼
EXAMPLE 4.2
Show that
Za
xm1 ða xÞn1 dx ¼ amþn1 bðm; nÞ:
0
Solution. Putting
x ¼ ay, we get
Za
xm1 ða xÞn1 dx
m
n
xn1 ½1ð1xÞð1xÞm1 dx
0
2
m
¼ 4
n
Z1
Z1
xn1 ð1xÞm1 dx
0
3
xn1 ð1xÞm dx5
0
m
¼ ½bðn;mÞbðn;mþ1Þ
n
m
m
¼ bðm;nÞ bðmþ1;nÞ:
n
n
1þ
m
m
bðm þ 1; nÞ ¼ bðm; nÞ
n
n
or
ðayÞ
m1 n1
a
ð1 yÞ
n1
:a dy
0
Z1
ðn þ mÞbðm þ 1; nÞ ¼ mbðm; nÞ
or
a
¼a
0
Z1
0
0
xn
mð1xÞm1 ð1Þ: dx
n
Thus,
Z1
¼
0
xn1 :xð1xÞm1 dx
ðayÞm1 ða ayÞn1 :a dy
¼
n
Z1
Z1
Z1
¼
n 1
mx
0
0
0
m1þn1þ1 m1
y
ð1 yÞ
n1
dy
bðm þ 1; nÞ
m
¼
:
bðm; nÞ
mþn
Z1
mþn1
y m1 ð1 yÞn1 dy
0
¼ amþn1 bðm; nÞ:
EXAMPLE 4.3
Show that
bðm þ 1; nÞ
m
¼
:
bðm; nÞ
mþn
4.3
Solution. We have
Z1
bðmþ1;nÞ ¼ xm ð1xÞn1 dx
xm1 þ xn1
dx ¼ 2bðm; nÞ:
ð1 þ xÞmþn
Solution. We know that 1
Z
bðm; nÞ ¼
n
EXAMPLE 4.4
Prove that
Z2
1 mþ1 nþ1
m
n
;
;
sin h cos hdh ¼ b
2
2
2
0
m > 1 and n > 1:
4.4
n
Engineering Mathematics-II
Solution. Put bx ¼ ay so that dx ¼ ab dy in the given
integral. This gives
Z1 aym1
Z1
xm1
a
b
dx
¼
: dy
ða þ bxÞmþn
ða þ ayÞmþn b
Solution. We have
Z2
sinm h cosn h dh
0
Z2
0
¼
sin
m1
h cos
n1
h: sin h cos h dh
0
Z2
¼
¼
0
Z1
1
an b m
0
sin
m1
hð1 sin hÞ
2
n1
2
sin h cos h dh:
ym1
1
dy ¼ n m bðm; nÞ;
ab
ð1 þ yÞmþn
using property (4) of beta function.
0
Putting sin2 h ¼ x so that 2sin h cos h dh ¼ dx, we
get
Z2
sinm h cosn h dh
0
Z1
Solution. We know [see property (2)] that
Z1
nþ1
1
mþ1
x ð1 xÞ dx ¼
x 2 1 ð1 xÞ 2 1 dx
2
0
0
1 mþ1 nþ1
¼ b
;
; m > 1 and n > 1:
2
2
2
1
¼
2
m1
2
EXAMPLE 4.7
Show that b m; 12 ¼ 22m1 bðm; mÞ.
n1
2
Z2
sin2m1 h cos2n1 h dh:
bðm; nÞ ¼ 2
ð5Þ
0
Putting n ¼
1
2,
we get
Z2
1
b m;
¼ 2 sin2m1 h dh:
2
EXAMPLE 4.5
Show that
0
bðm; nÞ ¼ bðm þ 1; nÞ þ bðm; n þ 1Þ:
Now, putting n ¼ m in (5), we have
Z2
Solution. By definition,
bðm; mÞ ¼ 2
bðm þ 1; nÞ þ bðm; n þ 1Þ
Z1
0
Z1
xm ð1 xÞn1 dx þ
¼
0
xm1 ð1 xÞn dx
0
Z2
0
m1
x
ð1 xÞ
n1
½x þ ð1 xÞdx
0
Z1
2m1
Z2 1
¼2
sin 2h
dh
2
0
xm1 ð1 xÞn1 dx ¼ bðm; nÞ:
¼
0
EXAMPLE 4.6
R1 xm1
Express ðaþbxÞ
mþn dx; m; n; a; b > 0 in terms of
0
ðsin h cos hÞ2m1 dh
¼2
Z1
¼
sin2m1 h cos2m1 h dh
beta function.
¼
1
22m1
Z2
:2
sin2m1 2h dh
0
¼
1
Z
sin2m1 d; ¼ 2h
22m1
0
ð6Þ
Beta and Gamma Functions
¼
1
22m1
Z2
:2
sin2m1 d
0
1
¼ 2m1 b m; ; using ð6Þ;
2
2
1
and so,
4.3
1
¼ 22m1 bðm; mÞ:
b m;
2
GAMMA FUNCTION
The gamma function is defined as the definite integral
Z1
ex xn1 dx; n > 0:
ðnÞ ¼
0
The gamma function is also known as Euler’s Integral
of Second Kind.
4.4
PROPERTIES OF GAMMA FUNCTION
1. We have
Z1
Z1
x n
n x 1
ðnþ1Þ ¼ e x dx ¼ ½x e 0 þn xn1 ex dx
0
0
Z1
¼n
ex xn1 dx ¼ nðnÞ:
ð4Þ ¼ 3:ð3Þ ¼ 3:2:1 ¼ 3!; and so on:
Moreover, (0) ¼ 1 and ( n) ¼ 1 if n > 0.
Also,
ðn þ 1Þ
ðn þ 1Þðn þ 1Þ
ðnÞ ¼
; n 6¼ 0 ¼
n
nðn þ 1Þ
ðn þ 2Þ
; n 6¼ 0 and n 6¼ 1
¼
nðn þ 1Þ
ðn þ 2Þðn þ 2Þ
¼
nðn þ 1Þðn þ 2Þ
ðn þ 3Þ
; n 6¼ 0; n 6¼ 1; and n 6¼ 2
¼
nðn þ 1Þðn þ 2Þ
ðn þ k þ 1Þ
; n 6¼ 0; n 6¼ 1;
¼
nðn þ 1Þðn þ 2Þ:::ðn þ kÞ
n 6¼ 2; and n 6¼ k:
Thus, (n) for n < 0 is defined, where k is a leastpositive integer such that n þ k þ 1 > 0.
4.5
RELATION BETWEEN BETA AND GAMMA
FUNCTIONS
We know that
Z1
ðmÞ ¼
Hence,
¼ nðn 1Þðn 2Þðn 2Þ
¼ nðn 1Þðn 2Þ:::3:2:1 ð1Þ
¼ n!ð1Þ:
But, by definition,
Z1
ð1Þ ¼
ex dx ¼ ½ex 1
0 ¼ 1:
et t m1 dt:
0
Putting t ¼ x2 so that dt ¼ 2xdx, we get
Z1
2
ðmÞ ¼ 2 ex x2m1 dx:
ð7Þ
0
Similarly, we can have
Z1
2
ðnÞ ¼ 2 ey y2n1 dy:
0
Therefore,
Z1
ðmÞðnÞ ¼ 4
e
x2 2m1
x
Z1
dx:
0
0
Hence,
ðn þ 1Þ ¼ n!; when n is a positive integer:
If we take n ¼ 0, then
0! ¼ ð1Þ ¼ 1;
4.5
and so, gamma function defines 0!
Further, from the relation (n þ 1) ¼ n(n), we
deduce that
ð2Þ ¼ 1:ð1Þ ¼ 1!;
ð3Þ ¼ 2:ð2Þ ¼ 2:1 ¼ 2!;
0
ðn þ 1Þ ¼ nðnÞ;
which is called the recurrence formula for (n).
2. Let n be a positive integer. By property (1), we
have
ðn þ 1Þ ¼ nðnÞ ¼ nðn 1Þðn 1Þ
n
2
0
Z1 Z1
¼4
0
ey y2n1 dy
0
eðx
2
þy2 Þ 2m1 2n1
x
y
dx dy:
4.6
n
Engineering Mathematics-II
Taking x ¼ r cos h and y ¼ r sin h, we have
dx dy ¼ rdh dr.
Therefore,
Z 2 Z1
2
er r2ðmþnÞ1 cos2m1 h
ðmÞðnÞ ¼ 4
0
0
sin2n1 h drdh
3
2
Z 2 Z1
2
¼ 2 42 er r2ðmþnÞ1 dr5
0
cos
0
2m1
h sin
" Z 2
¼ ðm þ nÞ 2
2n1
h dh
cos2m1 h
0
#
sin2n1 h dh ; usingð1Þ
¼ ðm þ nÞbðm; nÞ using property (7)
of beta function:
Hence,
EXAMPLE 4.9
R2 pffiffiffiffiffiffiffiffiffiffi
R2 pffiffiffiffiffiffiffiffiffiffi
Express the integrals
tan h dh and
cot h dh
0
in terms of gamma function.
ðmÞðnÞ
:
ðm þ nÞ
1
Putting m ¼ n ¼
2, we
get 1 1
2
1 1
¼ 2
b ;
2 2
ð1Þ
2
1
¼ :
2
bðm; nÞ ¼
1 1
¼b ;
2 2
Z1
¼
x21 ð1 xÞ21 dx
1
1
0
Z1
¼
0
2
12
:
¼
2
2
pffiffiffi
1
¼ :
2
Hence,
Solution. We know that
We know that (see Example 4.4)
Z2
1 mþ1 nþ1
sinm h cosn hdh ¼ b
;
2
2
2
0
mþ1 nþ1
1 2 2
:
¼
2 mþnþ2
2
Putting m ¼ n ¼ 0, we get
2
2
Z2
12
12
¼
:
dh ¼
2ð1Þ
2
0
ðmÞðnÞ
bðm; nÞ ¼
:
ðm þ nÞ
2
Second Method
Thus,
EXAMPLE 4.8
pffiffiffi
Show that 12 ¼ :
Thus,
1
2
"
!#1
1
x
dx
1
2
¼ qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
ffi ¼ sin
1
1 2
2
0
0
4 x2
¼ sin1 ð1Þ sin1 ð1Þ ¼ ¼ :
2
2
Hence,
pffiffiffi
1
¼ :
2
Z1
dx
pffiffiffipffiffiffiffiffiffiffiffiffiffi ¼
x 1x
Z1
0
dx
pffiffiffiffiffiffiffiffiffiffiffi
x x2
0
Solution. We have
1
Z 2 pffiffiffiffiffiffiffiffiffiffi
Z2
Z2
1
sin2 h
1
tan h dh ¼
dh ¼ sin2 h cos 2 h dh
1
2
cos h
0
0
0
1 1 2þ1
2þ1
2 2
34 14
1 1 ¼
¼
þ2
2ð1Þ
2 2 22
1
3
1
:
¼ 2
4
4
Similarly, we can show that
Z 2 pffiffiffiffiffiffiffiffiffiffi
1
3
1
:
cot h dh ¼ 2
4
4
0
Beta and Gamma Functions
EXAMPLE 4.10
Show that ðnÞð1 nÞ ¼ sinn ; 0 < n < 1:
Z2
(Euler’s Reflection Formula)
0
Solution. We know that
Z1
bðm; nÞ ¼
0
Also,
bðm; nÞ ¼
Therefore,
xn1
dx:
ð1 þ xÞmþn
ðmÞðnÞ
; m > 0 and n > 0:
ðm þ nÞ
ðmÞðnÞ
¼
ðm þ nÞ
Z1
n1
x
dx:
ð1 þ xÞmþn
0
Putting m ¼ 1n so that m > 0 implies n < 1, we get
Z1 n1
ðnÞð1 nÞ
x
dx
¼
ð1 þ xÞ
ð1Þ
n
pþ1 qþ1
2 2
sin h cos h dh ¼
:
2 pþqþ2
2
p
q
4.7
ð8Þ
Putting q ¼ 0 in (8), we get
1
pffiffiffi
Z2
pþ1
pþ1
2 2
2
:
sinp h dh ¼
pþ2 ¼ pþ2
2
2 2
2
0
Similarly, taking p ¼ 0, we get pffiffiffi
Z2
ðqþ1Þ
2
q
cos h dh ¼ qþ2
:
2
2
0
EXAMPLE 4.12
Show that ðmÞ m þ
1
2
¼
pffiffiffi
ð2mÞ
22m1
ðDuplication FormulaÞ:
0
or
Z1
ðnÞð1 nÞ ¼
xn1
dx ¼
; 0 < n < 1:
1þx
sin n
0
EXAMPLE 4.11
Show that
Z2
ðmÞðnÞ
:
sin2m1 h cos2n1 hdh ¼
2ðm þ nÞ
0
Hence, evaluate
R2
0
sinp h dh and
R2
cosp h dh:
0
Solution. We know that
Z2
2 sin2m1 h cos2n1 h dh ¼ bðm; nÞ
0
or
Z
2
1
sin2m1 h cos2n1 h dh ¼ :bðm; nÞ:
2
0
But, bðm; nÞ ¼ ðmÞðnÞ
ðmþnÞ . Therefore,
Z2
Solution. In Example
4.7, we have shown that
1
b m;
¼ 22m1 bðm; mÞ:
2
Converting into gamma function, we get
ðmÞ 12
ðmÞðmÞ
¼ 22m1
:
ð2mÞ
m þ 12
pffiffiffi
Since 12 ¼ p
;ffiffiffiwe get
ðmÞ
¼ 22m1
1
ð2mÞ
mþ2
or
pffiffiffi
1
¼ 2m1 ð2mÞ:
ðmÞ m þ
2
2
EXAMPLE 4.13
Show that
Z1
eax xn1 dx ¼
ðnÞ
;
an
0
where a and n are positive. Deduce that
R1 ax n1
e x cos bx dx ¼ rðnnÞ cos nh
(i)
0
ðmÞðnÞ
sin2m1 h cos2n1 h dh ¼
:
2ðm þ nÞ
0
If we put 2m 1 ¼ p and 2n 1 ¼ q, then this
result reduces to
(ii)
R1
0
eax xn1 sin bx dx ¼ rðnnÞ sin nh;
where r2 ¼ a2 þ b2 and h ¼ tan1 ba. Also evaluate
R1 ax
R1
e cos bx dx and eax sin bx dx:
0
0
4.8
n
Engineering Mathematics-II
Solution. Put ax ¼ z, so that adx ¼ dz, to get
Z1
Z1 n1
z
dz
ax n1
e x dx ¼
ez
:
a
a
0
EXAMPLE 4.14
Show that 0
1
¼ n
a
Z1
ez zn1 dz ¼
ð nÞ
: ð9Þ
an
pffiffiffi
ð2n þ 1Þ
:
2n
2 ðn þ 1Þ
pffiffiffi
Hence, deduce that 14 34 ¼ 2.
nþ
1
2
¼
0
Replacing a by a þ ib in (9), we get
Z1
ð nÞ
eðaþibÞx xn1 dx ¼
:
ða þ ibÞn
ð10Þ
Solution. We know that
Z2
bðm; nÞ ¼ 2 sin2m1 h cos2n1 h dh:
0
0
But as,
eðaþibÞx ¼ eax :eibx ¼ eax ðcos bx i sin bxÞ
and taking a ¼ r cos h and b ¼ r sin h, De-Moivre’s
Theorem implies
ða þ ibÞn ¼ ðr cos h þ ir sin hÞn
¼ rn ðcos h þ i sin hÞn
0
¼ rn ðcos nh þ i sin nhÞ:
1
Therefore, (10) reduces to
Z1
½eax ðcosbxisinbxÞxn1 dx
0
ð nÞ
ð nÞ
¼ n
¼ n ðcosnhþisinnhÞ1
r ðcosnhþisinnhÞ
r
ð nÞ
¼ n ðcosnhisinnhÞ:
r
Equating real- and imaginary parts on both sides, we get
Z1
ð nÞ
eax xn1 cos bx dx ¼ n cos nh
r
and
0
Z1
eax xn1 sin bx dx ¼
ð nÞ
sin nh:
rn
0
If we put n ¼ 1, then
Z2
ð 1Þ
r cos h
a
cos h ¼
eax cos bx dx ¼
¼ 2
r
r2
a þ b2
0
and
Z2
0
Therefore,
1
1
b nþ ;nþ
2
2
Z2 Z2
sin 2h 2n
2n
2n
dh
¼ 2 sin hcos h dh ¼ 2
2
eax sin bx dx ¼
ð 1Þ
r sin h
b
sin h ¼ 2 ¼ 2
:
r
r
a þ b2
¼
1
sin 2h dh ¼ 2n :
2
Z
sin2n d; ¼ 2h
2n
22n1
0
1
¼ 2n :2
2
0
Z2
Z2
0
" pffiffiffi#
2nþ1
2
sin d ¼ 2n1 2nþ2
2
2
2
2n
0
1
ðsee Example 2:11Þ
pffiffiffi n þ 12
:
¼ 2n
2 ðn þ 1Þ
Also,
n þ 12 n þ 12
1
1
b n þ ;n þ
¼
ð2n þ 1Þ
2
2
2
n þ 12
:
¼
ð2n þ 1Þ
From (11) and (12), we have
pffiffiffi
1
ð2n þ 1Þ
nþ
¼ 2n :
:
2
ðn þ 1Þ
2
Further, putting n ¼ 14, we have
pffiffiffi 3 rffiffiffi
3
¼ pffiffiffi : 25 ¼
4
2
2 4
Hence,
1
1
2 2
:
1
1
4 4
pffiffiffi
3
1
¼ 2:
4
4
ð11Þ
ð12Þ
Beta and Gamma Functions
EXAMPLE 4.15
Show that
Z 2 pffiffiffiffiffiffiffiffiffiffi
Z 2 pffiffiffiffiffiffiffiffiffiffi
1 1
3
¼ pffiffiffi :
tanh dh ¼
coth dh ¼ 2 4
4
2
0
0
0
0
0
0
0
EXAMPLE 4.16
Show that
Z1
1 p1
ðpÞ
q1
y
log
dy ¼ p ; p > 0; q > 0:
y
q
0
Solution. Putting log 1y ¼ x, we have 1y ¼ e x or y ¼ ex
x
and so, dy ¼ e
dx. Therefore,
1
Z
1 p1
yq1 log
dy
y
0
Z0
eðq1Þx x p1 ðex Þdx
¼
0
Solution. We have
Z2
Z 2 pffiffiffiffiffiffiffiffiffi
dh
sinh dh: pffiffiffiffiffiffiffiffiffi
sinh
0
But,
pffiffiffi
1
3
¼ 2 ðExample 2:14:Þ
4
4
Hence,
Z 2 pffiffiffiffiffiffiffiffiffiffi
Z 2 pffiffiffiffiffiffiffiffiffiffi
tan h dh ¼
cot h dh ¼ pffiffiffi :
2
Z2
¼
1
2
Z2
sin h dh:
0
1 sin 2 h dh:
¼
0
Z1
¼
0
1
¼ p
q
1
0
pffiffiffi 12þ1 pffiffiffi
2
34 14 :
¼ 1þ2 :
: ¼ :
2 12þ2 2 54 34 4
2
2þ1
2
2
2
ðsee Example 5:11Þ
14 14 ¼ 5 : ¼ 1 1 : ¼ :
4 4 4 4 4
EXAMPLE 4.18
Prove that
Z1 1 n1
log
dy ¼ ðnÞ; n > 0:
y
0
Solution. Putting log 1y ¼ x; that is, 1y ¼ e x or y ¼ ex,
we have dy ¼ exdx. Hence,
Z1 Z0
1 n1
log
dy ¼ xn1 ex dx
y
1
0
Z1
1
Z1
¼
eqx x p1 dx
et
Z1
0
ex xn1 dx
0
¼ ðnÞ; n > 0:
p1
t
dt
: ; putting qx ¼ t
q
q
et tp1 dt ¼
4.9
EXAMPLE 4.17
Show that
Z2
Z 2 pffiffiffiffiffiffiffiffiffiffi
dh
sin h dh: pffiffiffiffiffiffiffiffiffiffi ¼ :
sin h
0
Solution. In Example 4.9, we have proved that
Z 2 pffiffiffiffiffiffiffiffiffiffi
Z 2 pffiffiffiffiffiffiffiffiffiffi
1
1
3
:
tan h dh ¼
cot h dh ¼ 2
4
4
n
ðpÞ
:
qp
EXAMPLE 4.19
3
R1 Evaluate x4 log 1x dx:
0
Solution. Putting log 1x ¼ y, that is, 1x ¼ ey or x ¼ ey,
we have dx ¼ ey dy. Therefore,
4.10
n
Engineering Mathematics-II
Z1
1
x
x4 log
3
0
Solution. Putting x2 ¼ tan h, we have 2xdx ¼ sec2 h dh.
Therefore,
dx ¼ e4y :y3 :ey dy ¼ e5y :y3 dy Z1
Z4
dx
sec2 h
1
1
0
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dh
1
4
Z
1þx
1 þ tan2 h
2 tan h
3
dt
0
0
t t
: ; putting 5y ¼ t
¼ e :
Z4
Z4
125 5
2
1
sec
h
1
1
dh
0
pffiffiffiffiffiffiffiffiffiffi :
dh ¼
¼
1
1
Z1
2
sec
h
2
tan h
sin2 h cos2 h
1
1
6
0
0
t 3
¼
ð3Þ ¼
:
e :t dt ¼
Z4
Z4
625
625
625
1
dh
1
dh
0
¼
¼
2 ðsin h cos hÞ12 2 sin 2h12
Z0
Z1
0
EXAMPLE 4.20
Show that
Z1
x m ðlog xÞn dx ¼
0
1
¼ pffiffiffi
2
ð1Þn n!
ðm þ nÞ
:
nþ1
1
¼ pffiffiffi
2 2
y
Solution. Putting log x ¼ y, we have x ¼ e and so,
dx ¼ ey. When x ¼ 1, y ¼ 0 and when x ? 0,
y ? 1. Therefore,
2
¼ pffiffiffi
4 2
xm ðlogxÞn dx
¼
e
n
dh
1
pffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffi
2
sin 2h
Z2
0
0
d
pffiffiffiffiffiffiffiffiffiffi; ¼ 2h
2 sin sin 2 cos0 d
1
Z2
sin 2 cos0 d
1
R2
since bðm; nÞ ¼ 2 sin2m1 cos2n1 d:
0
my
Z2
0
1
1 1
¼ pffiffiffi b ; ;
4 2 4 2
Z1
Z0
0
2
0
Z4
y
Z1
ðyÞ ðe Þdy ¼ ð1Þ
n
0
ðmþ1Þy
e
:y dy
n
EXAMPLE 4.22
Show that
n
Z1
t
dt
n
Z 2 pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi
et :
¼ ð1Þ
; putting ðmþ1Þ y ¼ t
mþ1 mþ1
tan h þ sec h dh
1
0
0
¼
¼
Z1
ð1Þn
ðmþ1Þ
nþ1
0
ð1Þn n!
ðmþ1Þnþ1
et tn dt ¼
ðnþ1Þ
ðmþ1Þnþ1
0
" pffiffiffi #
1
1
3
¼ þ 3 :
2
4
4
4
:
Solution. We have
Z 2 pffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffi
tan h þ sec h dh
EXAMPLE 4.21
Prove that
Z1
0
ð1Þn
0
dx
1
1 1
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffi b ; :
1 þ x4 4 2 4 2
Z 2 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Z 2 pffiffiffiffiffiffiffiffiffiffi
tan h dh þ
sec h dh
¼
0
0
Beta and Gamma Functions
Z2
1
2
¼
sin h cos
1
2
Z2
h dh þ
0
¼
2
2
Let
sin h cos
0
1
2
h dh
I¼
Z
sin2 h cos 2 h dh þ
1
1
0
2
2
1 3 1
1 1 1
þ b ;
¼ b ;
2 4 4
2 2 4
" #
12 14
1 34 14
þ
¼
ð1Þ
2
34
#
" 12
1
1
3
þ 3
¼ 2
4
4
4
"
pffiffiffi #
1
1
3
¼ þ 3 :
2
4
4
4
4.6
Z
sin0 h cos 2 h dh
where D is the domain where X
0, Y
Y þ Z 1. Thus,
Z1 Z1Y
qþr
I ¼a
Y q1 Z r1 dZ dY
0
¼ aqþr
(The Dirichlet’s Theorem can be extended to a
finite number of variables).
Proof: Since x þ y þ z 1, we have y þ z 1 x ¼ a.
Therefore,
ZZZ
x p1 y q1 z r1 dx dy dz
V
Z
Z1 Z1x 1xy
xp1 yq1 zr1 dx dy dz
¼
x
0
0
2
p1 4
3
Za Zay
y
0
0
q1 r1
z
aqþr
r
Zr
r
1Y
dY
0
Z1
Y q1 ð1 Y Þr dY
0
V
x
Y q1
0
¼
0, and
0
Z1
Theorem 4.1 (Dirichlet). If V is the region, where
x Z Z0,
Z y 0, z 0, and x þ y þ z 1, then
ðpÞðqÞðrÞ
xp1 yq1 zr1 dx dy dz ¼
:
ðp þ q þ r þ 1Þ
0
ðaY Þq1 ðaZÞr1 :a2 dZ dY ;
I¼
D
0
The following theorems of Dirichlet and Liouville
are useful in evaluating multiple integrals.
Z1
0
Putting y ¼ aYZ and z ¼ aZ, this integral reduces to
1
DIRICHLET’S AND LIOUVILLE’S THEOREMS
¼
yq1 zr1 dz dy:
0
2
4.11
Za Zay
0
2
n
dz dy5dx:
aqþr
aqþr ðqÞðr þ 1Þ
¼
bðq; r þ 1Þ ¼
r
r ðq þ r þ 1Þ
ðqÞðrÞ
:
¼ aqþr
ðq þ r þ 1Þ
Hence, (1) yields
ZZZ
xp1 yq1 zr1 dx dy dz
V
Z1
ðqÞðrÞ p1 qþr
x a dx
ðqþr þ1Þ
0
Z
ðqÞðrÞ
x p1 ð1xÞqþr dx; since a ¼ 1x
¼
ðqþr þ1Þ
ðqÞðrÞ
¼
bðp; qþr þ1Þ
ðqþr þ1Þ
ðqÞðrÞ ðpÞðqþr þ1Þ ðpÞðqÞðrÞ
:
¼
:
¼
ðqþr þ1Þ ðpþqþr þ1Þ ðpþqþr þ1Þ
¼
Remark 4.1. If x þ y þ z h, then by putting
y
x
z
h ¼ X ; h ¼ Y ; and h ¼ Z, we have X þ Y þ Z h
h ¼ 1 and so, the Dirichlet’s Theorem takes the form
ZZZ
x p1 y q1 z r1 dx dy dz
V
ð1Þ
¼
ðpÞðqÞðrÞ pþqþr
:
:h
ðq þ r þ 1Þ
4.12
n
Engineering Mathematics-II
Theorem 4.2 (Liouville). If x, y, and z are all positive
such that
Z Z Zh1 < x þ y þ z < h2, then
f ðx þ y þ zÞxp1 yq1 zr1 dx dy dz
¼
ðpÞðqÞðrÞ
ðp þ q þ rÞ
Zh2
f ðhÞhpþqþr1 dh:
h1
(Proof, not provided here, is a slight modification of
the proof of Dirichlet’s Theorem).
EXAMPLE 4.23
RRR
Evaluate
x y z dx dy dz taken throughout the
ellipsoid
x 2 y2 z2
þ þ 1:
a2 b2 c 2
Solution. Put
x2
a2
¼ X;
1
2
y2
b2
¼ Y , and
z2
c2
1
2
¼ Z to get
1
x ¼ aX ; y ¼ bY ; and z ¼ cZ 2
Solution. We wish to evaluate
ZZZ
dx dy dz
under the condition ax þ by þ cz ¼ 1. Putting
y
x
z
a ¼ X ; b ¼ Y , and c ¼ Z, we get X þ Y þ Z ¼ 1.
Also dx ¼ adX, dy ¼ bdY, and dz ¼ cdZ. Therefore,
using Dirichlet’s Theorem, the required volume of
the tetrahedron is
ZZZ
V¼
dx dy dz
ZZZ
¼
abc dX dY dZ
ZZZ
¼ abc
X 11 Y 11 Z 11 dX dY dZ
ð1Þð1Þð1Þ
ð1 þ 1 þ 1 þ 1Þ
abc
abc abc
¼
¼
:
¼
ð4Þ
3!
6
¼ abc
and
a2
b2
c2
xdx ¼ dX ; ydy ¼ dY ; and zdz ¼ dZ:
2
2
2
The condition, under this substitution, becomes
EXAMPLE 4.25
RRR l1 m1 n1
z
dx dy dz, where
Evaluate
x y
p x >q 0,
y > 0, and z > 0 under the condition ax þ by þ
zr
c 1.
X þ Y þ Z 1:
Therefore,
Z Z Z for the first quadrant,
xyz dx dy dz
ZZZ
¼
ðxdxÞðydyÞðzdzÞ
Z Z Z 2 2 2 a
b
c
¼
dX
dY
dZ
2
2
2
Z
Z
Z
a2 b2 c 2
X 11 Y 11 Z 11 dX dY dZ
¼
8
a2 b2 c2 ð1Þð1Þð1Þ
;
¼
8
ð1 þ 1 þ 1 þ 1Þ
by Dirichlet’s Theorem
a2 b2 c 2 1
a2 b2 c 2 a2 b2 c 2
:
¼
:
¼
¼
8
8:3!
48
ð4Þ
Therefore, value
for the whole of the
2 2of2 the integral
2 2 2
b c
ellipsoid is 8 a 48
¼ a b6 c :
EXAMPLE 4.24
The plane ax þ by þ cz ¼ 1 meets the axes in A, B, and
C. Find the volume of the tetrahedron OABC.
Solution. Put
x p
a
¼ X;
y q
b
¼ Y ; and
z r
c
¼ Z so that
a 1
dx ¼ X p1 dX ;
p
b 1
dy ¼ Y q1 dY ; and
q
c 1
dz ¼ Z r1 dZ:
r
Therefore,
ZZZ
xl1 ym1 zn1 dx dy dz
ZZZ 1 m1 1 n1
1 l1
bY q
cZ r
¼
aX p
abc 1p1 1q1 11
X Y Z r dX dY dZ
pqr
ZZZ
l
m
al bm c n
n
¼
X p1 Y q 1 Z r 1 dX dY dZ
pqr
l
m
n
al bm c n p q r
:
: ¼
pqr 1 þ l þ m þ n
p
q
r
Beta and Gamma Functions
EXAMPLE 4.26
R R m1 n1
y 2 dx dy over the
Show that
x
positive
2
m n
octant of the ellipse ax2 þ by2 ¼ 1 is a2nb b m2 ; n2 þ 1 :
2
¼
2
Solution. Putting ax2 ¼ X and by2 ¼ Y , we get x ¼
pffiffiffiffi
pffiffiffiffi
1
a X and y ¼ b Y and dx ¼ a2 X 2 dX and dy ¼ b2
1
Y 2 dY . Therefore,
ZZ
xm1 yn1 dxdy
ZZ
n1 a
1 b
1
m1
¼
am1 X 2 bn1 Y 2 X 2 Y 2 dXdY
2
2
Z Z
am bn
m
n
X 2 1 Y 21 dX dY
¼
4
am bn m2 n2
am bn m2 n2 n2
¼ 4n : mþn
: ¼
4 1 þ m2 þ n2
2 þ1
2
am bn m2 n2 þ 1
am bn m n
:
b
;
þ
1
:
¼
¼
2n
2n
2 2
mþn
2 þ1
EXAMPLE 4.28
R a pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Evaluate 0 x4 a2 x2 dx using Gamma function.
Solution. Putting x p
¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
a sin h, we get
ffi dx ¼ a cos h dh
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
2
2
2
and a x ¼ a a sin h ¼ a cos h. When
x ¼ 0; h ¼ 0 and when x ¼ a; h ¼ 2. Therefore
Za pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
I ¼ x4 a2 x2 dx
0
Z2
a4 sin4 hða cos hÞða cos hÞdh
¼
0
Z2
MISCELLANEOUS EXAMPLES
EXAMPLE 4.27
R1
Evaluate 0
Since
using Beta and Gamma functions.
0
Z1
2
1
t 3
1
t31
dt ¼
1 2 dt
1þt
6
ð1 þ tÞ3þ3
0
0
1
1 2
;
; since bðm; nÞ
¼ b
6
3 3
Z1
xm1
¼
dx
ð1 þ xÞmþn
0
1 13 23
1 13 23
¼
¼
ð1Þ
6 13 þ 23
6
1
1
2
¼ 6
3
3
1
¼
6
Z1
sin4 h cos2 h dh:
6
0
sin h cosn h dh ¼ 12
m
n1
ðmþ1
2 Þ ð 2 Þ
ðmþnþ2
2 Þ
, we have
5 3
2 2
2 ð4Þ
pffiffiffi pffiffiffi
a6 32 : 12 : 12 a6
:
¼
¼
2
32
3!
I ¼ a6
1
Solution. Putting x6 ¼ t, we have x ¼ t6 and so
5
dx ¼ 16 t 6 . When x ¼ 0; t ¼ 0 and when x ¼ 1;
t ¼ 1. Therefore
Z1 1
Z1
x dx
1
t6 5
t 6 dt
¼
1þt
1 þ x6 6
0
R2
0
x dx
1þx6
4.13
1
2
;
using
ðnÞ
1
6 sin 3
n
ð0 < nÞ:
¼
sin n
¼a
4.7.
n
EXAMPLE 4.29
R 1 xn
... ðn1Þ
ffidx ¼ 2:4:
Show that 0 pffiffiffiffiffiffiffi
1:3: 5: ... n , where n is odd
1x2
integer.
Solution. Putting x ¼ sin h, we have dx ¼ cos h dh.
Therefore
Z1
0
xn
pffiffiffiffiffiffiffiffiffiffiffiffi dx ¼
1 x2
Z2
0
sinn h
cos h dh
cos h
Z2
¼
sinn h dh
0
pffiffiffi
nþ1
2 nþ2
(see Example 4.11):
¼
2 2
4.14
n
Engineering Mathematics-II
Since n is odd, we take n ¼ 2m þ 1. Therefore
Z1
xn
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx
1 x2
0
pffiffiffi
ðm þ 1Þ ¼
2 m þ 32 2
pffiffiffi
mðm 1Þðm 2Þ . . . 3:2:1
¼
1
1
3
3 1
1 2
ðm þ 2Þðm 2Þðm 2Þ . . . 2 2 ð2Þ
2mð2m 2Þð2m 4Þ . . . 6:4:2
ð2m þ 1Þð2m 1Þð2m 3Þ . . . 3:1
ðn 1Þðn 3Þðn 5Þ . . . 6:4:2
; ðn oddÞ
¼
nðn 2Þ . . . 5:3:1
¼
which was to be established.
2. Show that
Z1
pffiffiffi x3
xe dx ¼
pffiffiffi
:
3
0
3. Show that
Z1
x2n1 eax dx ¼
2
ðnÞ
:
2an
0
4. Show that
Z1 a
x
ða þ 1Þ
dx ¼
; if a > 1:
ax
ðlog aÞaþ1
0
5. Show that
Z2
Solution. Putting ax ¼
Therefore
Z1
xm e a
2 2
x
pffiffi
1
z, we have a dx ¼ 12 z 2 dz.
8
:
77
0
6. Show that
Zb
1
dx ¼
2a
0
5
sin3 x cos2 xdx ¼
EXAMPLE 4.30
R1
2 2
Show that 0 xm e a x dx ¼ 2a1mþ1 mþ1
2 .
Z1
pffiffim
z
1
z
e
z 2 dz
a
ðxaÞm ðbxÞn dx ¼ ðbaÞmþnþ1 bðmþ1;nþ1Þ:
a
0
¼
¼
1
2amþ1
1
2amþ1
Z1
0
Z1
0
1
¼ mþ1 2a
EXERCISES
1. Show that
:
(i) bð2:5; 1:5Þ ¼ 16
9 7
5
:
¼
(ii) b 2 ; 2
pffiffiffi
3 1 2048
(iii) 4 4 ¼ 2:
ez z
m1
2
ez z
7. Prove that
dz
mþ1
2 1
Z1
dz
mþ1
:
2
xn ea x dx ¼
2 2
1
nþ1
; n > 1:
2anþ1
2
0
8. Show that
Z1
0
x2 dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi :
1 x4
Z1
0
dx
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffi:
4
4 2
1þx
9. Prove that
Z1
0
xdx
1 2 1
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ b ; :
1 x5 5 5 2
Beta and Gamma Functions
ZZZ
10. Show that
Z1
0
Hint : Mass ¼
pffiffiffi
1n
dx
:
pffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
:
n 1n þ 12
1 xn
(ii)
(iii)
R1 0
R1
1 n1
log x
ex dx
0
12. Show that
2
R1
0
13. Show that
R1
0
dx
(ii)
(iii) 12
R1
0
R1
0
x3 ex dx ¼ 19 2
ey yn1 dy ¼ ðnÞ.
et t 2 dt ¼ 12 1
1
3
1
2
.
x2n1 eax dx ¼ ðnÞ
2an .
2
R1 pffiffiffi x2
R1 x2
xe dx: epffiffix dx ¼ 2p ffiffi2.
0
0
Put
x y
þ
a b
x
y
z
¼ X ; ¼ Y ; and ¼ Z and proceed:
a
b
c
b c
.
Ans. ma720
17. Show that the volume of the solid bounded
pffiffi
by the coordinate planes and the surface ax þ
qffiffi pffiffi
y
z
abc
bþ
c ¼ 1 is 90 .
2 2 2
2
.
14. Show that yb(x þ 1, y) ¼ xb(x, y þ 1).
15. Show that
4.15
z
þ 1
c
ZZZ
¼
mxyz dx dy dz
11. Express the integrals in terms of gamma
function:
R1
Ans. (i) ðnÞ
(i) xp1 ekx dx; k > 0
kn .
0
dx dy dz; 0 n
16. The plane ax þ by þ cz ¼ 1 meets the axes in A, B,
and C, respectively. Find the mass of the
tetrahedron OABC if the density at any point is
¼ m xyz.
2
2
18. Find the volume of the ellipsoid ax2 þ by2 þ cz2 ¼ 1.
Z Z Z
Hint : V ¼ 8
dxdydz:
x2
y2
z2
¼ X ; 2 ¼ Y ; and 2 ¼ Z;
2
a
b
c
and use Dirichlet’s Theorem to get
4
V¼
abc:
3
Put
19. Show that the entire volume of the solid
x23 y23 z23
4
a þ b þ c ¼ 1 is 35 abc.
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5
Laplace Transform
The study of Laplace transform is essential for
engineers and scientists because these transforms provide easy and powerful means of solving differential and integral equations. The
Laplace transforms directly provides the solution
of differential equations with given boundary
values without finding the general solution
first. A Laplace transform is an extension of
the continuous-time Fourier transform motivated
by the fact that this transform can be used to a
wider class of signals than the Fourier transform
can. In fact, Fourier transform does not converge for many signals whereas the Laplace
transform does. Fourier transform is not applicable to initial-value problems whereas Laplace
transform is applicable. Also some functions
like sinusoidal functions and polynomials do
not have Fourier transform in the usual sense
without the introduction of generalized functions. Causal functions (which assume zero
value for t < 0) are best handled by Laplace
transforms.
5.1
DEFINITION AND EXAMPLES OF LAPLACE
TRANSFORM
Definition 5.1. Let f (t) be a function of t defined for
t > 0. Then the one-sided Laplace transform (or
merely Laplace transform) of f(t), denoted by
L{ f (t)} or F(s), is defined by
Z1
Lf f ðtÞg ¼ FðsÞ ¼ est f ðtÞdt;
s 2 R or C;
0
provided that the integral converges for some value
of s.
The defining integral is called the Laplace
integral.
Definition 5.2. The two-sided Laplace transform of
a function f (t) is defined by
Z1
est f ðtÞdt
Lf f ðtÞg ¼
1
for all values of s, real or complex for which the
integral converges.
The defining integral in this case is called twosided Laplace integral.
The symbol L, which transforms f (t) into F(s) is
called Laplace operator. Thus the Laplace transform
of a function exists only if the Laplace integral converges. The following theorem provides sufficient
conditions for the existence of Laplace transform.
Theorem 5.1. Let f (t) be piecewise continuous in
every finite interval 0 t N and be of exponential
order c for t > N. Then Laplace transform of f (t)
exists for all s > c.
Proof: Since f (t) is piecewise continuous on every
finite interval [0, N] and est is also piecewise
continuous on [0, N] for N > 0, it follows that e–st
f (t) is integrable on [0, N]. For any positive number
N, we have
ZN
Z1
Z1
st
st
e f ðtÞ dt ¼ e f ðtÞdt þ est f ðtÞdt:
0
0
N
By the above arguments, the first integral on the
right exists. Further, since f (t) is of exponential
order c for t > N, there exists constant M such that
| f (t)| M ect for t 0 and so
Z1
Z1
Z1
st
st
j e f ðtÞdtj je f ðtÞjdt jest f ðtÞjdt
N
N
Z1
0
0
est M ect dt ¼
M
:
sc
Thus the Laplace transform L { f (t)} exists for s > c.
5.2
n
Engineering Mathematics-II
Remark 5.1. The conditions of the Theorem 5.1 are
sufficient but not necessary for the existence of
Laplace transform of a function. Thus Laplace
transform may exist even if these conditions are not
satisfied. For example, f (t) ¼ t–1/2 does not satisfy
these conditions but its Laplace transform does
exist (see Example 5.8).
EXAMPLE 5.1
Find Laplace transform of unit step function f
defined by f (t) ¼ 1, t 0.
Solution. By definition of Laplace transform, we have
ZT
Z1
st
est dt
Lf f ðtÞg ¼ e dt ¼ lim
es
e2s
ð1 es Þ þ 2
ð1 es Þ
s
s
e3s
ð1 es Þ þ . . .
þ3
s
es
ð1 es Þ½1 þ 2es þ 3e2s þ . . .
¼
s
es
1
ð1 es Þ
¼
s
ð1 es Þ2
es
1
¼
:
¼
sð1 es Þ sðes 1Þ
¼
EXAMPLE 5.4
Find Laplace transform of f (t) ¼eat, t
0.
Solution. By the definition of Laplace transform,
ZT
Z1
we have
st at
est T
1 est 1
e
e
dt
¼
lim
eðsaÞt dt
Lf
f
ðtÞg
¼
¼ if s > 0:
¼ lim
¼ lim
T !1
T !1 s 0
T !1
s
s
0
0
T!1
0
0
EXAMPLE 5.2
Find the Laplace transform of the unit ramp function
f defined by f (t) ¼ t, t 0.
Solution. Using integration by parts, we get
ZT
Z1
st
Lf f ðtÞg ¼ t e dt ¼ lim t est dt
T !0
0
0
( )
est T est T
¼ lim
t
2
T !1
s 0
s 0
1 esT T esT
1
¼ 2 if s > 0:
¼ lim 2 2 T !1 s
s
s
s
EXAMPLE 5.3
Find L{ f (t)}, where f (t) ¼ [t], t > 0.
Solution. We have
Z1
Lf f ðtÞg ¼ est f ðtÞ dt
¼
e
0
st
T !1
eðsaÞ
ðs aÞ
T
1 eðsaÞT
T!1
sa
¼ lim
0
1
; if s > a:
¼
sa
The result of this example holds for complex numbers also.
EXAMPLE 5.5
Find Laplace transforms of f (t) ¼ sin at and g (t) ¼
cos at.
Solution. Since
Z
eat ða sin bt b cos btÞ
;
eat sin bt dt ¼
a 2 þ b2
and
Z
eat ða cos bt þ b sin btÞ
eat cos bt dt ¼
;
a2 þ b 2
we have
Z1
Lfsin atg ¼
0
Z1
¼ lim
Z2
ð0Þ dt þ
e
1
st
Z3
dt þ
0
e
2
est sin at dt
st
st 2
st 3
st 4
e þ 2 e þ 3 e þ . . .
¼ s 1
s 2
s 3
2dt þ . . .
ZT
¼ lim
T !1
0
¼ lim
T !1
est sin at dt
est ðs sin at a cos atÞ
s 2 þ a2
T
0
Laplace Transform
a
est ðs sin a T þ a cos aT Þ
¼ lim 2
T !1 s þ a2
s 2 þ a2
a
if s > 0;
¼ 2
s þ a2
and
Z1
Lfcos atg ¼
est cos at dt
ZT
¼ lim
T!1
1
snþ1
Z
0
eu uðnþ1Þ1 du ¼
est cos at dt
est ðs cos at þ a sin atÞ T
¼ lim
T!1
s 2 þ a2
0
s
est ðs cos aT a sin aT Þ
¼ lim 2
T!1 s þ a2
s 2 þ a2
s
if s > 0:
¼ 2
s þ a2
EXAMPLE 5.6
Find the Laplace transforms of f (t) ¼ sinh at and
g (t) ¼ cosh at.
Solution. Since sinh at ¼ e
at
eat
,
2
we have
Z1 at at e e
¼ est
dt
2
eat eat
2
Lfsinh atg ¼ L
Solution. Putting st ¼ u, we have
Z1 n
Z1
u du
:
Lf f ðtÞg ¼ est tn dt ¼ eu
s
s
¼
0
0
Z1
Z1
1
1
st at
¼
e e dt est eat dt
2
2
0
0
1
1
¼ Lfeat g Lfeat g
2
2
1 1
1
a
¼ 2
¼
; s > jaj:
2 sa sþa
s a2
at
Again, since cosh at ¼ e þe
; proceeding as
2
above, we have
1
1
Lfcosh atg ¼ Lfeat g þ Lfeat g
2
2
1 1
1
þ
¼
2 sa sþa
s
; s > jaj:
¼ 2
s a2
at
5.3
EXAMPLE 5.7
Find Laplace transform of f (t) ¼ tn, where n is a
positive integer.
0
0
n
ðn þ 1Þ
snþ1
for s > 0 and n þ 1
0;
by the definition of gamma function. Since n is
positive integer, G(n þ 1) ¼ n! and so
n!
Lftn g ¼ nþ1 :
s
Remark 5.2. Integrating the defining formula for
Laplace transform of tn by parts, we have
Z1
n
Lft g ¼ tn est dt
0
tn est
¼
s
1
0
n
þ
s
Z1
tn1 est dt:
0
The integral on the right exists and the lower limit
can be used in the first term if n 1. Since s > 0,
the exponent in the first term goes to zero as t tends
to infinity. Thus, we obtain a general recurrence
formula,
n
Lftn g ¼ Lftn1 g; n 1:
s
Hence, by induction, we get the sequence
1
ðby Example 5:1Þ
Lft0 g ¼ Lf1g ¼
s
1
1
Lftg ¼ Lft0 g ¼ 2
s
s
2
Lftg ¼ 3
s
......
......
n!
Lftn g ¼ nþ1 :
s
EXAMPLE 5.8
Find the Laplace transform of f ðtÞ ¼ t1=2 .
5.4
n
Engineering Mathematics-II
Solution. The condition n þ 1 > 0 of Example 5.7 is
satisfied and so
pffiffiffi rffiffiffi
ð1=2Þ
1=2
Lft
g ¼ 1=2 ¼ 1=2 ¼
:
s
s
s
It may be mentioned here that the function f ðtÞ ¼
t1=2 does not satisfy the conditions of Theorem 5.1,
even then the Laplace transform of this function
exists. Thus the conditions of Theorem 5.1 are
sufficient but not necessary for the existence of
Laplace transform of a given function.
EXAMPLE 5.9
Find the Laplace transform of the function f defined by
t for 0 t 3
f ðtÞ ¼
0 for t > 3:
Solution. The graph of the function f is shown in the
Figure 5.1.
3
f (t )
t
0
3
Figure 5.1
Using integration by parts, we have
Z1
Z3
st
Lf f ðtÞg ¼ e f ðtÞ dt ¼ est tdt
0
¼ t:
0
st 3
e
s
0
Z3
Solution. Integration
by parts yields
Z1
pffiffiffiffi st
tn e dt
Lf f ðtÞg ¼
0 pffiffiffiffi
Z1 pffiffiffiffiffiffiffiffi
1
n
tn est
þ
¼
tn2 est dt:
s
2s
0
0
If n 1, the lower limit can be used in the first term
on the right and thus the integral exists. Thus
pffiffiffiffi
n pffiffiffiffiffiffiffiffi
Lf tn g ¼ Lf tn2 g; n 1 and odd:
2s
Thus we obtain a sequence of formulas given below:
pffiffiffi
1
ðExample 5:8Þ
L pffi ¼ pffiffi
s
t
pffiffiffi
pffi
Lf tg ¼ pffiffiffiffi
2 p
s3ffiffiffi
pffiffiffiffi
3
Lf t3 g ¼ pffiffiffiffi
4 s5
......
......
pffiffiffi
pffiffiffiffi
ðn þ 1Þ! n
p
Lf t g ¼
ffiffiffiffiffiffiffiffiffiffiffi :
2ðnþ1Þ nþ1
sðnþ2Þ
2 !
EXAMPLE 5.11
pffiffi
Find the Laplace transform of erfð zÞ and erf(z).
Solution. Recall that the error function is defined by
the integral
Zz
2
2
erfðzÞ ¼ pffiffiffi et dt;
0
where the variable z may be real or complex.
The graph of erf(z), where z is real is shown in
the Figure 5.2.
est
dt
s
0
3
1 est 3
¼ e3s þ
s
s s 0
3 3s 1 3s
¼ e 2 ½e 1
s
s
1
3
¼ 2 ½1 e3s e3s for s > 0:
s
s
EXAMPLE 5.10
Find the Laplace
pffiffiffiffi transform of the function f defined
by f ðtÞ ¼ tn ; n 1 and odd integer.
1
erf(z)
z
−2
−1
1
2
−1
Figure 5.2 The Error Function
Laplace Transform
pffiffi
Let us find Laplace transform of erf ð zÞ: Using
2
series expansion of e–t , we have
pffiffi
Lferfð zÞg
9
8
pffi
>
>
=
< 2 Zt
4
6
t
t
2
1 t þ þ . . . dt
¼ L pffiffiffi
>
>
2! 3!
;
: x = t
s=2
and so
1
1
1:3
1:3:5
¼ 3=2 5=2 þ
þ ...
7=2
s
2s
2:4s
2:4:6 s9=2
1
1 1:3 1 1:3:5 1
1 þ
þ ...
2s 2:4 s2 2:4:6 s3
s3=2
1
1 1=2
¼ 3=2 1 þ
s
s
¼
s > 1:
We now find the Laplace transform of erf (z).
We have
Z1
LferfðzÞg ¼ est erfðzÞ dz
0
Zz
Z1
2
st 2
p
ffiffiffi
ex dx dz
¼ e
0
0
Changing the order of integration (Figure 5.3),
we get
Z1 Z1
Z1
2
2
2
x2
st
LferfðzÞg ¼ pffiffiffi e
e dt dx ¼ pffiffiffi eðx þsxÞ dx
s 0
x
Z1
s 2
2 s2 =4
¼ pffiffiffi e
eðxþ2Þ dx;
s 0
2 2
because x2 þ sx ¼ x þ 2s s4 .
Figure 5.3
Taking u ¼ x þ 2s , we have
Z1
2 s2 =4
2
LferfðzÞg ¼ pffiffiffi e
eu du
s ð7=2Þ
ð9=2Þ
þ ...
7=2
5:2! s
7:3! s9=2
1
¼ pffiffiffiffiffiffiffiffiffiffiffi ;
s sþ1
t
0
2
t3=2 t5=2 t7=2
¼ pffiffiffi L t1=2 þ
þ ...
3
5:2! 7:3!
þ
5.5
x
0
2 ð3=2Þ ð5=2Þ
¼ pffiffiffi
s3=2
3s5=2
n
0
s
2
2
LferfðzÞg ¼ pffiffiffi es =4 erf c
; s > 0:
2
s EXAMPLE 5.12
pffi
Find the Laplace transform of f ðtÞ ¼ sin t.
pffi
Solution. The series expansion for sin sin t is
pffi
t3=2 t5=2 t7=2
þ
þ ...
sin t ¼ t1=2 3!
5!
7!
1
1
X
ð1Þn tnþ2
¼
ð2n þ 1Þ!
n¼0
Therefore
1
X
pffi
ð1Þn
1
Lfsin tg ¼
Lftnþ2 g
ð2n
þ
1Þ!
n¼0
1
X
ð1Þn n þ 32
¼
ð2n þ 1Þ! snþ32
n¼0
1
X
ð1Þn
ð2n þ 2Þ! pffiffiffi 1
: 2nþ2
: nþ3
ð2n þ 1Þ! 2
ðn þ 1Þ!
s 2
n¼0
p
ffiffiffi
1
X
ð1Þn ð2n þ 1Þ!
: 2nþ1 : nþ3
¼
ð2n
þ
1Þ!
n!
2
s 2
n¼0
rffiffiffi 1
1 X ð1Þn 1 n
¼
2s s n¼0 n!
4s
rffiffiffi
1 1
¼
e 4s ; s > 0:
2s s
¼
5.6
n
Engineering Mathematics-II
EXAMPLE 5.13
Find the Laplace transform of the pulse of unit
height and duration T.
EXAMPLE 5.16
Find the Laplace transform of a function defined by
t
for 0 < t < a
f ðtÞ ¼ a
1 for t > a:
Solution. The pulse of unit height and duration T is
defined by
1 for 0 < t < T
f ðtÞ ¼
0 for T < t:
Therefore,
ZT
1 esT
:
Lf f ðtÞg ¼ est dt ¼
s
Solution. Integrating by parts,
Za
Z1
t st
Lf f ðtÞg ¼
e dt þ est dt
a
a
0
esa esa 1 esa 1 esa
¼
¼
þ
:
s
s
as2
as2
0
EXAMPLE 5.14
Find the Laplace transform of sinusoidal (sine) pulse.
Solution. Sinusoidal pulse is defined by
for 0 < t < =a
for =a < t:
sin at
0
f ðtÞ ¼
Therefore, by the definition of Laplace transform
we have
Z1
að1 þ es=a Þ
Lf f ðtÞg ¼
sin at est dt ¼
:
s 2 þ a2
0
The denominator of L{ f (t)} here is zero at s ¼ ± ia.
But, since e±ia ¼ –1, the numerator also becomes
zero. Thus L{ f (t)} have no poles and is an entire
function.
EXAMPLE 5.15
Find the Laplace transform of triangular pulse of
duration T.
EXAMPLE 5.17
Find Laplace transform of f defined by
et for 0 < t < 1
f ðtÞ ¼
0 for t > 1:
Solution. By definition of Laplace transform, we have
Z1
Z1
e1s 1
t st
:
Lf f ðtÞg ¼ e :s dt ¼ eðsþ1Þt dt ¼
1s
0
EXAMPLE 5.18
Find the Laplace transform of a function f defined by
sin t for 0 < t < f ðtÞ ¼
0
for t > :
Solution. The integration by parts yields
Z
Z
1
st
est cos tdt
Lf f ðtÞg ¼ e sin t dt ¼ 0 þ
s
0
¼
Solution. The triangular
defined by
8
< T2 t
f ðtÞ ¼ 2 T2 t
:
0
pulse of duration T is
for 0 < t < T=2
for T=2 < t < T
for T < t:
By the definition of Laplace transform, we have
ZT =2
ZT 2
2
st
Lf f ðtÞg ¼
t e dt þ
2 t est dt
T
T
0
¼
T=2
sT =2
2 1 2e
þe
s2
T
sT
:
0
¼
0
1 est
1
cos t þ
s s
s
0
1
1
½1 þ es 2
s2
s
Z
est
sin t dt
s
0
Z
est sin t dt:
0
Thus,
2
s þ1
1
Lf f ðtÞg ¼ 2 ½1 þ es ;
s2
s
which yields
Lf f ðtÞg ¼
1 þ es
:
s2 þ 1
Laplace Transform
EXAMPLE 5.19
Find the Laplace transform of the function fe
defined by
1
for 0 t e
fe ðtÞ ¼ e
0 for t > e;
where e > 0. Deduce the Laplace transform of the
Dirac delta function.
Solution. The graph of the function fe is shown in the
Figure 5.4.
fε ( t )
1
ε
ε
t
0
Figure 5.4
Z
þ
e!0
LfðtÞg ¼ limf fe ðtÞg ¼ 1:
e 0
The function (t) is called the Dirac delta function
or unit impulse function having the properties
Z1
ðtÞ ¼ 0; t 6¼ 0; and ðtÞ dt ¼ 1:
0
EXAMPLE 5.20
Find the Laplace transform of Heaviside’s unit step
function defined by
1 for t > a
Hðt aÞ ¼
0 for t < a:
Solution. The Heaviside’s unit step function is also
known as delayed unit step function and occurs in
the electrical systems. It delays the output until t ¼ a
and then assumes a constant value of one unit. Its
graph is shown in the Figure 5.5.
H(t − a)
t
a
0
Figure 5.5
est fe ðtÞ dt
Ze
e
st
1 ese
:
dt ¼
se
0
Further we note that
1 ese
lim Lf fe ðtÞg ¼ lim
e!0
e!0
se
2 2
1 ð1 se þ s 2!e . . .Þ
¼ lim
e!0
se se
¼ lim 1 þ . . . ¼ 1:
e!0
2
Also, we observe from the definition
of fe that
lim fe ðtÞdoes not exist and so L lim fe ðtÞ is not
e!0
1
0
e
1
¼
e
5.7
defined. Even
then itis useful to define a function as ðtÞ ¼ lim fe ðtÞ such that
We observe that as e ! 0, the height of the rectangle
increases indefinitely and the width decreases in
such a way that its area is always equals to 1.
The Laplace transform of fe is given by
Z1
Ze
st
Lf fe ðtÞg ¼ e fe ðtÞ dt ¼ est fe ðtÞ dt
0
n
e!0
The Laplace transform of Heaviside’s unit function
is given by
Z1
Z1
st
LfHðt aÞg ¼ e Hðt aÞ dt ¼ est dt
a
0
ZT
¼ lim
est dt ¼ lim
T !1
T!1
a
est
s
T
¼
a
esa
:
s
EXAMPLE 5.21
Find the Laplace transform of rectangle function
defined by
1 for a < t < b
gðtÞ ¼
0 otherwise:
5.8
n
Engineering Mathematics-II
Solution. The graph of this function is shown in the
Figure 5.6. Clearly, this function can be expressed
in terms of Heaviside’s unit function as
gðtÞ ¼ Hðt aÞ Hðt bÞ:
Proof: Using the definition of Laplace transform and
the linearity property of integral, we have
Lfc1 f1 ðtÞ þ c2 f2 ðtÞg
Z1
¼ est ½c1 f1 ðtÞ þ c2 f2 ðtÞ dt
0
Z1
Z1
st
¼ c1 e f1 ðtÞ dt þ c2 est f2 ðtÞ dt
g(t )
0
1
¼ c1 F1 ðsÞ þ c2 F2 ðsÞ:
t
0
a
b
EXAMPLE 5.22
Find the Laplace transform of f (t) ¼ sin2 3t.
Solution. Since
Figure 5.6
Further if a ¼ 0, then g becomes pulse of unit height
and duration b (Example 5.13). The Laplace transform of rectangle function g is given by
LfgðtÞg ¼ LfHðt ag LfHðt bÞg
¼
esa esb
s
s
esa esb
:
¼
s
5.2
0
PROPERTIES OF LAPLACE TRANSFORMS
While studying the following properties of Laplace
transforms, we assume that the Laplace transforms
of the given functions exist.
Theorem 5.2. (Linearity of the Laplace Transform). If
c1 and c2 are arbitrary constants (real or complex)
and f1(t) and f2(t) are functions with Laplace
transforms F1(s) and F2(s), respectively, then
Lfc1 f1 ðtÞ þ c2 f2 ðtÞg ¼ c1 Lf f1 ðtÞg þ c2 Lf f2 ðtÞg
¼ c1 F1 ðsÞ þ c2 F2 ðsÞ:
Thus L is a linear operator.
sin2 3t ¼
we have
1 cos 6t 1 1
¼ cos 6t;
2
2 2
1 1
1
1
Lfsin2 3tg ¼ L cos6t ¼ Lf1g Lfcos 6tg
2 2
2
2
1 1
1
s
¼
; s>0
2 s
2 s 2 þ 62
¼
1 1
s
18
¼ 2
; s > 0:
2
2 s s þ 36
sðs þ 36Þ
EXAMPLE 5.23
Find the Laplace transform of f (t) ¼ e4t þ e2t þ t3 þ
sin2t.
2t
Solution. Since sin2 t ¼ 1cos
; by Theorem 5.2,
2
we have
Lf f ðtÞg ¼ Lfe4t þ e2t þ t3 þ sin2 tg
1
¼ Lfe4t g þ Lfe2t g þ Lft3 g þ Lf1g
2
1
Lfcos 2tg
2
1
1
6
1
s
þ
þ þ ¼
s 4 s 2 s4 2s 2ðs2 þ 4Þ
1
1
6
2
; s > 0:
¼
þ
þ 4þ 2
s4 s2 s
sðs þ 4Þ
EXAMPLE 5.24
Find Laplace transform of f (t) ¼ sin3 2t.
Laplace Transform
Solution. Since sin 3t ¼ 3 sin t – 4 sin3 t, we have
3
1
sin3 t ¼ sin t þ sin 3t
4
4
and so
5.9
we have by linearity of the operator L,
Lf f ðtÞg ¼ cos Lfsin vtg þ sin Lfcos vtg
v
s
þ sin 2
; s>0
¼ cos 2
s þ v2
s þ v2
3
1
sin3 2t ¼ sin 2t sin 6t:
4
4
Hence, by linearity of L, we get
3
1
Lf f ðtÞg ¼ Lfsin 2tg Lfsin 6tg
4
4
3
2
1
6
; s>0
¼
2
2
4 s þ4
4 s þ 36
¼
n
3
1
1
2 s2 þ 4 s2 þ 36
¼
1
s2 þ v2
ðv cos þ s sin Þ; s > 0:
EXAMPLE 5.27
Determine Laplace transform of the square wave
function f defined by
f ðtÞ ¼ HðtÞ 2Hðt aÞ þ 2Hðt 2aÞ
2Hðt 3aÞ þ . . .
48
:
¼ 2
ðs þ 4Þ ðs2 þ 36Þ
Solution. We note that
EXAMPLE 5.25
Find the Laplace transform of f (t) ¼ sin at sin bt.
Solution. We have
1
f ðtÞ ¼ ð2 sin at sin btÞ
2
1
¼ ½cosðat btÞ cosðat þ btÞ
2
1
1
¼ cosða bÞt cosða þ bÞt:
2
2
f ðtÞ ¼ HðtÞ 2Hðt aÞ ¼ 1 2ð0Þ ¼ 1; 0 < t < a
f ðtÞ ¼ HðtÞ 2Hðt aÞ þ 2Hðt 2aÞ
¼ 1 2ð1Þ þ 2ð0Þ ¼ 1; 0 < a < t < 2a
and so on. Thus the graph of the function is as shown
in the Figure 5.7.
By linearity of Laplace operator, we have
Lf f ðtÞg ¼ LfHðtÞ 2LfHðt aÞg
þ 2LfHðt 2aÞg 2LfHðt 3aÞg þ . . .
Therefore, using linearity, we have
1
1
Lff ðtÞg ¼ Lfcosða bÞtg Lfcosða þ bÞtg
2
2
"
# "
#
1
s
1
s
; s>0
¼
2 s2 þ ða bÞ2 2 s2 þ ða þ bÞ2
1
esa
e2sa
e3sa
þ2
2
þ ...
¼ 2
s
s
s
s
f(t)
¼
2abs
ðs2 þ ða bÞ2 Þ ðs2 þ ða þ bÞ2 Þ
; s > 0:
1
EXAMPLE 5.26
Find the Laplace transform of f (t) ¼ sin(ot þ ),
t 0.
0
t
a
2a
−1
Solution. Since
sinðvt þ Þ ¼ sin vt cos þ cos vt sin ;
Figure 5.7
3a
4a
5.10
n
Engineering Mathematics-II
Also
1
¼ ½1 2esa f1 esa þ e2sa . . .g
s
1
1
¼ 1 2esa
s
1 þ esa
sa
sa 1 1 esa
1 e 2 e 2
¼
¼
sa
sa
s 1 þ esa
s e 2 þ e 2
1
sa
¼ tanh
:
s
2
Hence
Lf f ðtÞg ¼ Lfeat cos bt þ i eat sin btg
¼ Lfeat cos btg þ i L feat sin btg
ðby linearity of LÞ
Solution. Since
ðsin t cos tÞ2 ¼ sin2 t þ cos2 t 2 sin t cos t
¼ 1 sin 2t;
1
2
Lf f ðtÞg ¼ Lf1g Lfsin 2tg ¼ 2
s s þ4
s2 2s þ 4
; s > 0:
¼
sðs2 þ 4Þ
ðs aÞ2 þ b2
:
Comparing real and imaginary parts, we have
sa
Lfeat cos btg ¼
ðs aÞ2 þ b2
and
b
Lfeat sin btg ¼
:
ðs aÞ2 þ b2
sin u
u
du:
Solution. Using series expansion of sin u, we have
Zt
Zt
sin u
1
u3 u5 u7
du ¼
u þ þ . . . du
u
u
3! 5! 7!
0
0
Zt
¼
1
u2 u4 u6
þ þ . . . du
3! 5! 7!
0
¼t
t3
t5
t7
þ
þ ...
3:3! 5:5! 7:7!
Therefore,
EXAMPLE 5.30
Find Laplace transform of eat cos bt and eat sin bt,
where a and b are real.
Solution. Let f (t) =e(a+ib)t. Then (see Example 5.4)
1
1
¼
L f f ðtÞg ¼
s ða þ ibÞ s a ib
1
ðs aÞ þ ib
:
¼
ðs aÞ ib ðs aÞ þ ib
ðs aÞ2 þ b2
ðs aÞ þ ib
0
Solution. By linearity of L, we have
pffi
1
Lf f ðtÞg ¼ 2Lf1g þ Lf tg þ L pffi
t
1
pffiffiffi
þ
1
2
¼ þ 21þ1 þ
s
s
s2
rffiffiffi
pffiffiffi
2
¼ þ 3=2 þ
; s > 0:
s 2s
s
ðs aÞ þ ib
Lfeat cos btg þ i L feat sin btg ¼
EXAMPLE 5.31
Rt
Find the Laplace transform of f ðtÞ ¼
EXAMPLE 5.29
pffi
Find Laplace transform of f ðtÞ ¼ 2 þ t þ p1ffit ; t > 0.
¼
ð2Þ
Thus, by (1) and (2), we have
EXAMPLE 5.28
Find the Laplace transform of f (t) = (sin t cos t)2,
t 0.
we have
eðaþibÞt ¼ eat ½cos bt þ i sin bt
¼ eat cos bt þ i eat sin bt
ð1Þ
t3
t5
t7
þ
þ ...
3:3! 5:5! 7:7!
1
1 3!
1 5!
1 7!
¼ 2
: þ
: :
s
3:3! s4 5:5! s6 7:7! s8
þ . . . ðby linearity of LÞ
Lf f ðtÞg ¼ L t 1
1
1
1
4 þ 6 8 þ ...
2
s
3s
5s
7s
"
#
1 1 ð1=sÞ3 ð1=sÞ5 ð1=sÞ7
þ
þ ...
¼
3
5
7
s s
¼
1
1
¼ tan1 :
s
s
Laplace Transform
EXAMPLE 5.32
Find Laplace transform f (t) = cosh at cos at.
Proof: We are given that
Lf f ðtÞg ¼ FðsÞ ¼
Solution. By linearity of the Laplace operator, we have
L f f ðtÞg ¼ Lfcosh at cos atg
¼ Lfcosh atg Lfcos atg
s
s
; s > jaj
¼ 2
s a2 s 2 þ a2
s 3 þ a2 s s 3 þ a2 s
2a2 s
¼
¼
:
s 4 a4
s 4 a4
EXAMPLE 5.33
Find the Laplace transform of Bessel’s function of
order zero.
Solution. Recall that Bessel’s function of order zero
is defined by
t2
t4
t6
J0 ðtÞ ¼ 1 2 þ 2 2 2 2 2 þ . . .
2
2 :4
2 :4 :6
Therefore,
t2
t4
t6
LfJ0 ðtÞg ¼ L 1 2 þ 2 2 2 2 2 þ...
2 2 :4 2 :4 :6
¼ Lf1g
1
1
Lft2 gþ 2 2 Lft4 g
22
2 :4
1
2 2 2 Lft6 gþ...
2 :4 :6
1 1 2!
1 4!
1 6!
¼ 2 3 þ 2 2 5 2 2 2 7 þ...
s 2 s 2 4 s 2 :4 :6 s
1
1 1
1:3 1
¼ 1
þ
s
2 s2
2:4 s4
1:3:5 1
þ...
2:4:6 s6
"
#
1
1 1=2
ðusing binomial theoremÞ
1þ 2
¼
s
s
1
¼ pffiffiffiffiffiffiffiffiffiffiffi :
2
s þ1
Theorem 5.3. [First Shifting (Translation) Property].
If f (t) is a function of t for t > 0 and L{f (t)} = F(s),
then
Lfe f ðtÞg ¼ Fðs aÞ:
at
n
5.11
Z1
est f ðtÞ dt:
0
By the definition of Laplace transform, we have
Z1
Z1
at
st at
L fe f ðtÞg ¼ e ðe f ðtÞÞ dt ¼ eðsaÞ f ðtÞ dt
0
0
¼ Fðs aÞ:
EXAMPLE 5.34
Find the Laplace transform of g(t) = et sin2 t.
Solution. We have (see Example 5.23)
2
:
Lfsin2 tg ¼ FðsÞ ¼ 2
sðs þ 4Þ
Therefore, using first shifting property, we get
LfgðtÞg ¼ Fðs aÞ
2
¼
; since a ¼ 1:
ðs þ 1Þ ½ðs þ 1Þ2 þ 4
2
:
¼
2
ðs þ 1Þ ðs þ 2s þ 5Þ
EXAMPLE 5.35
Find Laplace transform of g(t) = t3e3t.
Solution. Since
3! 6
¼ ;
s4 s4
Lft3 g ¼ FðsÞ ¼
we have by shifting property,
LfgðtÞg ¼ Lfe3t :t3 g ¼ Fðs aÞ ¼
6
ðs þ 3Þ4
;
since a ¼ 3:
EXAMPLE 5.36
Using first-shifting property, find Laplace transforms of t sin at and t cos at.
Solution. Since Lftg ¼ s12 ; we have
Lft eiat g ¼ Lft cos atg þ iLft sin atg
¼ Fðs aÞ ¼
¼
1
ðs iaÞ
ðs2 a2 Þ þ ið2asÞ
ðs2 þ a2 Þ2
:
¼
2
ðs þ iaÞ2
½ðs iaÞ ðs þ iaÞ2
5.12
n
Engineering Mathematics-II
Equating real and imaginary parts, we have
Lft cos atg ¼
s 2 a2
ðs2 þ a2 Þ2
and
Lft sin atg ¼
2as
ðs2
þ a2 Þ 2
:
EXAMPLE 5.37
Find the Laplace transform of f (t) = eat cosh bt.
Therefore, by first shifting theorem, we have
e3t e3t
cos2 t
Lf f ðtÞg ¼ Lfsinh 3t cos2 tg ¼ L
2
1
1
¼ Lfe3t cos2 tg Lfe3t cos2 tg
2
2
ðby linearity of LÞ
"
#
1
ðs 3Þ2 þ 2
¼
2 ðs 3Þ½ðs 3Þ2 þ 4
"
#
1
ðs þ 3Þ2 þ 2
2 ðs þ 3Þ ½ðs þ 3Þ2 þ 4
Solution. Since
s
;
Lfcosh btg ¼ 2
s b2
the shifting property yields
s > jbj;
Lfeat cosh btg ¼ Fðs aÞ
sa
¼
; s > jbj þ a:
ðs aÞ2 b2
EXAMPLE 5.38
Find Laplace transform of f (t) = e3t (2cos 5t +
3 sin 5t).
EXAMPLE 5.40
Find the Laplace transform of cosh at sin bt.
Solution. Let F(s) be Laplace transform of f (t), t > 0
and let
gðtÞ ¼ f ðtÞ cosh at:
Then
LfgðtÞg ¼ L½ f ðtÞ cosh at ¼ L
Solution. Since
Lf2 cos 5t 3 sin 5tg
¼ 2Lfcos 5tg 3Lfsin 5tg
2s
35
2s 15
¼
¼ FðsÞ;
¼ 2
s þ 25 s2 þ 25 s2 þ 25
therefore, shifting property yields
Lf f ðtÞg ¼ Fðs aÞ with a ¼ 3
¼
1
s2 6s þ 11
2 ðs 3Þ ðs2 6s þ 13Þ
s2 þ 6s þ 11
:
ðs þ 3Þ ðs2 þ 6x þ 13Þ
¼
2ðs þ 3Þ 15
ðs þ 3Þ2 þ 25
¼
2s 9
:
s2 þ 6s þ 34
EXAMPLE 5.39
Find Laplace transform of f (t) = sinh 3t cos2 t.
2t
: Therefore,
Solution. We know that cos2 t ¼ 1þcos
2
1
1
Lfcos2 tg ¼ Lf1g þ Lfcos 2tg
2
2
1 1
s
s2 þ 2
¼ 2
; s > 0:
þ 2
¼
2 s s þ4
sðs þ 4Þ
eat þ eat
f ðtÞ
2
1
1
¼ Lðeat f ðtÞÞ þ Lðeat f ðtÞÞ
2
2
1
¼ ½Fðs aÞ þ Fðs þ aÞ
2
ðuse of first shifting theoremÞ:
b
We take f (t) = sin bt. Then FðsÞ ¼ s2 þb
2 and,
therefore, using above result, we have
Lfðcosh atÞ sin btg
"
#
1
b
b
þ
:
¼
2 ðs aÞ2 þ b2 ðs þ aÞ2 þ b2
EXAMPLE 5.41
Find the Laplace transform of f (t) = cosh 4t sin 6t.
Solution. Taking a = 4, b = 6 in Example 5.40, we get
Lfcosh 4t sin 6tg ¼
6ðs2 þ 52Þ
:
s4 þ 40s2 þ 2704
Laplace Transform
Theorem 5.4. (Second Shifting Property). Let F(s) be
the Laplace transform of f(t), t > 0 and let g be a
function defined by
f ðt aÞ for t > a
0
for t < a:
gðtÞ ¼
Then
0
¼
e
Z1
gðtÞdt þ
est gðtÞ dt
a
0
Z1
¼0þ
Using Heaviside’s unit step function H, this function can be expressed as
f ðtÞ ¼ Hðt 3Þ sin t:
To use second-shift theorem, we first write sin t as
sin t ¼ sinðt 3 þ 3Þ
Then
Proof: Using the substitution t a = u, we have
Z1
LfgðtÞg ¼ est gðtÞ dt
st
Lf f ðtÞg ¼ LfHðt 3Þ sinðt 3Þ cos 3g
þ LfHðt 3Þ cosðt 3Þ sin 3g
¼ cos 3e3s Lfsin tg þ sin 3e3s Lfcos tg
¼ cos 3 e3s
¼
est f ðt aÞ dt
0
Z1
as
¼e
esu f ðuÞ du ¼ eas FðsÞ:
0
EXAMPLE 5.42
Find the Laplace transform of the function f
defined by
8
2
2
>
< cos t for t >
3
3
f ðtÞ ¼
2
>
:
0
for t < :
3
s
Solution. We know that Lfcos tg ¼ s2 þ1
; s > 0.
Therefore, by second shifting property,
¼e
1
s
þ sin 3e3s 2
þ1
s þ1
e3s
ðcos 3 þ s sin 3Þ:
þ1
s2
se 3
; s > 0:
Lfcos tg ¼ 2
s þ1
f ðtÞ ¼
EXAMPLE 5.43
Find the Laplace transform of the sine function
switched on at time t = 3.
Solution. The given function is defined by
sin t for t 3
f ðtÞ ¼
0
for t < 3 :
ðt 1Þ2
0
for t 1
for 0 t < 1:
Solution. This function is just the function g (t) = t2
delayed by 1 unit of time and its graph is shown in
Figure 5.8.
f (t )
t
0
2s
Lf f ðtÞg
s2
EXAMPLE 5.44
Find the Laplace transform of the function f
defined by
a
Z1
¼ esðuþaÞ f ðuÞ du
2s
3
5.13
¼ sinðt 3Þ cos 3 þ cosðt 3Þ sin 3:
LfgðtÞg ¼ eas FðsÞ:
Za
n
1
Figure 5.8
Therefore, by second shift property, we have
2es
Lf f ðtÞg ¼ es Lft2 g ¼ 3 ; ReðsÞ > 0:
s
EXAMPLE 5.45
Find Laplace transform of the function f defined by
f ðtÞ ¼
ðt 4Þ5
0
for t > 4
for t < 4:
5.14
n
Engineering Mathematics-II
Solution. Using second shift property, we have
5!
e4s
Lf f ðtÞg ¼ e4s ½Lft5 g ¼ e4s : 6 ¼ 120 6 :
s
s
Then the Laplace transform of f 0 exists and is given by
Lf f 0 ðtÞg ¼ sFðsÞ f ð0Þ;
Theorem 5.5. (Change of Scale Property). If F(s) is
the Laplace transform of f (t) for t > 0, then for any
positive constant a,
1 s
Lf f ðatÞg ¼ F
:
a a
Proof: The existence of the Laplace transform is
established by Theorem 5.1. Further, integrating by
parts, we have
ZT
Z1
0
st 0
Lf f ðtÞg ¼ e f ðtÞdt ¼ lim est f 0 ðtÞdt
Proof: We are given that
where F(s) is the Laplace transform of f.
T !1
Z1
FðsÞ ¼ Lf f ðtÞg ¼ est f ðtÞ dt:
0
¼ lim
0
T!1:
Taking u = at, we have
Z1
Z1
du
st
Lf f ðatÞg ¼ e f ðatÞdt ¼ esu=a f ðuÞ
a
0
;
0
0
EXAMPLE 5.46
Find the Laplace of f (t) = cos 6t.
Theorem 5.6. (Laplace Transform of Derivatives).
Let f be a function such that
(a) f is continuous for all t, 0 t N
(b) f is of exponential order c for t > N
(c) f 0 is sectionally continuous for 0 t N.
est f ðtÞdt
9
=
Z1
¼ s est f ðtÞdt f ð0Þ
0
Solution. Let J0(t) be Bessel’s function of order zero.
1 ffi
: Therefore, by
By Example 5.33, LfJ0 ðtÞg ¼ pffiffiffiffiffiffiffi
s2 þ1
change of scale property,
1 s 1
1
1
LfJ0 ðatÞg ¼ F
¼ : qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi :
2 þ a2
a a
a
2
s
ðs=aÞ þ 1
½est f ðtÞT0 þ s
ZT
9
8
ZT
=
<
¼ lim ½esT f ðT Þ f ð0Þ þ s est f ðtÞdt
T!1:
;
0
EXAMPLE 5.47
Using change of scale property, find the Laplace
transform J0(at).
0
0
Z1
1
1 s
esu=a f ðuÞ du ¼ F
¼
:
a
a a
s
; the change of scale
Solution. Since Lfcos tg ¼ s2 þ1
property implies
!
1
s=6
L fcos 6tg ¼
6 ðs=6Þ2 þ 1
1
s
s
¼ 2
:
¼
6 6½ðs2 =36Þ þ 1
s þ 36
8
<
¼ sFðsÞ f ð0Þ;
the last but one step being the consequence of
the fact that f is of exponential order and so
lim esT f ðT Þ = 0 for s > c.
T !1
EXAMPLE 5.48
Find Laplace transform of g(t) = sin at cos at.
Solution. Let f (t) = sin2 at. Then
f 0 ðtÞ ¼ 2a sin at cos at:
Since
Lf f 0 ðtÞg ¼ sFðsÞ f ð0Þ;
we have
Lf2a sin at cos atg
¼ sLfsin2 atg 0 ¼ sLfsin2 atg
2a2
(see Example 5:22Þ:
þ 4a2 Þ
Hence
a
:
Lfsin at cos atg ¼ 2
ðs þ 4a2 Þ
EXAMPLE 5.49
Using Laplace transform of cos bt, find the Laplace
transform of sin bt.
¼
sðs2
n
Laplace Transform
Solution. We want to find L{sin bt} from L{cos bt}.
So, let f(t) = cos bt. Then f 0 (t) = b sin bt and so
Taking g(t) = f 0 (t), we have
Lf f 00 ðtÞ g ¼ sLf f 0 ðtÞg f 0 ð0Þ
0
Lf f ðtÞg ¼ sFðsÞ f ð0Þ ¼ sLfcos btg 1
s
s2
¼s 2
1
¼
1
s 2 þ b2
s þ b2
¼
Thus
Hence
b2
:
þ b2
s2
b2
Lfb sin btg ¼ 2
:
s þ b2
b
Lfsin btg ¼ 2
:
s þ b2
EXAMPLE 5.50
Find Laplace transform of Bessel’s function of
order 1.
Solution. Let J1(t) be Bessel’s function of order 1.
We know that
d n
ft Jn ðtÞg ¼ tn Jn1 ðtÞ:
dt
If n = 0, we have
J 00 ðtÞ ¼ J1 ðtÞ ¼ J 1 ðtÞ:
Hence
LfJ1 ðtÞg ¼ LfJ00 ðtÞg ¼ ½sLfJ 0 ðtÞ J0 ð0Þg
s
¼ pffiffiffiffiffiffiffiffiffiffiffiffi 1
2
s þ1
s
¼ 1 pffiffiffiffiffiffiffiffiffiffiffiffi (see Example 5.33Þ
2
s þ1
pffiffiffiffiffiffiffiffiffiffiffiffi
2
s þ1s
¼ pffiffiffiffiffiffiffiffiffiffiffiffi :
s2 þ 1
Theorem 5.7. If L{ f(t)} = F(s), then
Lf f 00 ðtÞ ¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ
if f(t) and f 0 (t) are continuous for 0 t N and of
exponential order for t > N whereas f@(t) is sectionally continuous for 0 t N.
Proof: By Theorem 5.6, we have
Lfg0 ðtÞg ¼ s GðsÞ gð0Þ:
5.15
¼ s½sFðsÞ f ð0Þ f 0 ð0Þ
¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ:
EXAMPLE 5.51
Using Theorem 5.7, find L{sin at}, t
0.
Solution. Let f(t) = sin at. Then
f 0 ðtÞ ¼ a cos at;
f 00 ðtÞ ¼ a2 sin at:
By Theorem 5.7,
Lf f 00 ðtÞg ¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ
and so
Lfa2 sin atg ¼ s2 Lfsin atg a
which yields
ðs2 þ a2 ÞLfsin atg ¼ a
and hence
Lfsin atg ¼
s2
a
; s > 0:
þ a2
EXAMPLE 5.52
Using Laplace transform of derivatives, find
L{t cos at}.
Solution. Let f (t) = t cos at. Then
f 0 ðtÞ ¼ cos at at sin at
f 00 ðtÞ ¼ 2a sin at a2 t cos at:
But
Lf f 00 ðtÞg ¼ s2 Lf f ðtÞg sf ð0Þ f 0 ð0Þ
¼ s2 Lf f ðtÞg 1
and so
Lf2a sin at a2 t cos tg ¼ s2 Lft cos atg 1;
that is,
ðs2 þ a2 ÞLft cos atg ¼ 2aLfsin atg þ 1
a
þ1
¼ 2a 2
s þ a2
s 2 a2
¼ 2
;
s þ a2
5.16
n
Engineering Mathematics-II
and so
Lft cos atg ¼
s 2 a2
ðs2 þ a2 Þ2
Therefore
Lfn!g ¼ sn Lftn g;
:
which gives
Theorem 5.7. can be generalized to higher order
derivatives in the form of the following result:
Theorem 5.8. Let L{ f (t)} = F(s). Then
Lf f ðnÞ ðtÞg ¼ sn FðsÞ sn1 f ð0Þ sn2 f 0 ð0Þ
. . . sf ðn2Þ ð0Þ f ðn1Þ ð0Þ;
if f(t), f 0 (t),. . ., f (n1)(t) are continuous for 0 t N
and of exponential order for t > N whereas f (n)(t) is
piecewise continuous for 0 t N.
Proof: We shall prove our result using mathematical
induction. By Theorems 5.6 and 5.7, we have
Lf f 0 ðtÞg ¼ sFðsÞ f ð0Þ;
Lf f 00 ðtÞg ¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ:
Lftn g ¼
Lfn!g n! Lf1g
n!
¼
¼ nþ1 :
sn
sn
s
Theorem 5.9. (Multiplication by tn). If L{ f(t)} = F(s),
then
d
Lftf ðtÞg ¼ FðsÞ;
ds
and in general
dn
Lftn f ðtÞg ¼ ð1Þn n FðsÞ:
ds
Proof: By definition of Laplace transform,
Z1
FðsÞ ¼ est f ðtÞ dt:
0
Thus the theorem is true for f 0 (t) and f@(t). Suppose
that the result is true for f (n) (t). Then
Lf f ðnÞ ðtÞ ¼ sn FðsÞ sn1 f ð0Þ . . . f ðn1Þ ð0Þ
Then application of Theorem 5.6 yields
Lf f ðnþ1Þ ðtÞg
Then, by Leibnitz-rule for differentiating under the
integral sign, we have
Z1
Z1
dF
d
d st
st
e f ðtÞ dt ¼
¼
ðe f ðtÞ dtÞ
ds ds
ds
0
¼ s½sn FðsÞ sn1 f ð0Þ . . . f ðn1Þ ð0Þ f ðnÞ ð0Þ
¼ snþ1 FðsÞ sn f ð0Þ . . . sf ðn1Þ ð0Þ þ f ðnÞ ð0Þ;
0
EXAMPLE 5.53
Using Theorem 5.8, find L{tn}.
Solution. We have f (t) = tn. Therefore,
f 0 ðtÞ ¼ ntn1 ; f 00 ðtÞ ¼ nðn 1Þtn2 ; . . . ; f ðnÞ ðtÞ ¼ n!
Now use of Theorem 5.8 yields
Lf f ðnÞ ðtÞg ¼ Lfn!g ¼ sn Lftn g
sn1 f ð0Þ . . . f ðn1Þ ð0Þ
But
f ð0Þ ¼ f 0 ð0Þ ¼ f 00 ð0Þ ¼ . . . ¼ f ðn1Þ ð0Þ ¼ 0:
0
¼ Lftf ðtÞg
th
which shows that the result holds for (n + 1)
derivative also. Hence by mathematical induction,
the result holds.
0
Z1
Z1
st
¼ t e f ðtÞ dt ¼ est ðt f ðtÞÞ dt
and so
d
FðsÞ:
ds
Thus the theorem is true for n = 1. To obtain the
general form, we use mathematical induction. So,
assume that the result is true for n = m. Thus
dm
Lftm f ðtÞg ¼ ð1Þm m FðsÞ ¼ ð1Þm FðmÞ ðsÞ:
ds
Therefore,
d
½Lðtm f ðtÞ ¼ ð1Þm F ðmþ1Þ ðsÞ;
ds
that is,
Z1
d
est tm f ðtÞ dt ¼ ð1Þm Fðmþ1Þ ðsÞ;
ds
Lftf ðtÞg ¼ 0
n
Laplace Transform
which, on using Leibnitz rule, yields
Z1
est tmþ1 f ðtÞ dt ¼ ð1Þm F ðmþ1Þ ðsÞ;
In a similar way, we can show that
Z1
t e2t cos t dt ¼ Lft cos tg with s ¼ 2
0
0
and so
5.17
"
¼
Lftmþ1 f ðtÞg ¼ ð1Þmþ1 F ðmþ1Þ ðsÞ:
Hence, the result follows by mathematical induction.
#
s2 1
ðs2
þ 1Þ
¼
2
s¼2
3
:
25
EXAMPLE 5.57
R1 2
Evaluate the integral I ¼ ex dx.
EXAMPLE 5.54
Find Laplace transform of f (t) = t sin2 t.
0
Solution. Let f(t) = t sin t. We know that Lfsin tg ¼
2
sðs2 þ4Þ : Therefore
"
#
d
2
3s2 þ 4
2
Lft sin tg ¼ :
¼2
ds sðs2 þ 4Þ
s2 ðs2 þ 4Þ2
2
2
EXAMPLE 5.55
Find Laplace transform of f(t) = tet cosh t.
2
Solution. Putting t = x , we get
Z1
1
1
I¼
et t1=2 dt ¼ Lft1=2 g with s ¼ 1
2
2
0
1 12 þ 1
¼ :
with s ¼ 1
2
s1=2 pffiffiffi
1
1
:
¼ ¼
2
2
2
Solution. We know that
Lfcosh tg ¼
Therefore,
Lft cosh tg ¼ s
:
s2 1
EXAMPLE 5.58
Find Laplace transform of f (t) ¼e–2t t cos t.
d s s2 þ 1
:
¼
ds s2 1
ðs2 1Þ2
Then, by Theorem 5.3, we have
Lfet t cosh tg ¼
ðs þ 1Þ2 þ 1
2
ððs þ 1Þ 1Þ
2
¼
s2 þ 2s þ 2
ðs2 þ 2sÞ2
:
EXAMPLE 5.56
R1
Using Theorem 5.9, evaluate t e2t sin t dt and
0
R1 2t
te cos t dt.
0
Solution. We know that L{sin t} = s2 1þ1.
Therefore, by Theorem 5.9, we have
d
1
2s
¼
:
Lft sin tg ¼ 2
2
ds s þ 1
ðs þ 1Þ2
1
R
But e2t ðt sin tÞdt is the Laplace transform of
s
Solution. We know that Lfcos tg ¼ s2 þ1
: Therefore,
by Theorem 5.9, we have
d
s
s2 1
¼
:
Lft cos tg ¼ ds s2 þ 1
ðs2 þ 1Þ2
Now using first shifting property, we have
Lfe2t t cos tg ¼
ðs þ 2Þ2 1
2
ððs þ 2Þ þ 1Þ
2
¼
s2 þ 4s þ 3
ðs2 þ 4s þ 5Þ2
EXAMPLE 5.59
Find the Laplace transform of f (t) ¼ t2 e–2t cos t.
1
Solution. As in Example 5.58, Lft cos tg ¼ ðss2 þ1Þ
2 :
Therefore,
!
d
s2 1
2
Lft cos tg ¼ :
ds ðs2 þ 1Þ2
2
0
t sin t with s = 2. Hence
"
Z1
2t
e ðt sin tÞdt ¼ Lft sin tg ¼
0
2s
ðs2 þ 1Þ2
#
Then using first-shifting property, we have
s¼2
4
¼ :
25
:
Lft2 e2t cos tg ¼ 2
s3 þ 10s2 þ 25s þ 22
ðs2 þ 4s þ 5Þ3
!
:
5.18
n
Engineering Mathematics-II
Solution. By linearity of L, we have
Lfcos 2t cos 3tg ¼ Lfcos 2tg Lfcos 3tg
s
s
:
¼ 2
1
Solution. Since Lfeat g ¼ sþa
; we have
s þ 4 s2 þ 9
n
Therefore, by Theorem 5.10, we get
1
n!
n d
n at
¼ ð1Þ2n
:
Lft e g ¼ ð1Þ n
nþ1
cos 2t cos 3t
ds s þ a
ðs þ aÞ
L
t
1
Z1
Z
Theorem 5.10. (Division by t). If L{ f (t)} ¼ F(s), then
u
u
du du
¼
Z1
2
2
u þ4
u þ9
f ðtÞ
s
s
L
¼ FðuÞ du;
t
1
1
1
2
s
¼ ½logðu þ 4Þs ½logðu2 þ 9Þ1
s
2
2
f ðtÞ
provided lim t exists.
1
2
t!0
1
u þ4
¼ log 2
f ðtÞ
2
u þ9 s
Proof: Put gðtÞ ¼ t : So, f (t) ¼ t g(t) and
1
u2 þ 4
1
s2 þ 4
Lf f ðtÞg ¼ LftgðtÞg
log 2
¼ lim log 2
2 u!1
u þ9
2
s þ9
d
¼ LfgðtÞg; by Theorem 5:9
2
1
1 þ ð4=u Þ
1
s2 þ 4
ds
¼ log lim
log
dG
u!1 1 þ ð9=u2 Þ
2
2
s2 þ 9
¼
:
2
2
ds
1
s þ9 1
s þ9
¼ 0 þ log 2
¼ log 2
:
Then integration yields
2
s þ4 2
s þ4
s
1
Z
Z
GðsÞ ¼ FðuÞ du ¼ FðuÞ du;
EXAMPLE 5.62
EXAMPLE 5.60
Find Laplace transform of f (t) ¼ tn eat.
1
that is,
L
f ðtÞ
t
Find the Laplace transform of f ðtÞ ¼ e
s
Z1
¼
FðuÞ du:
s
Remark 5.3. By Theorem 5.10, we have
Z1
Z1
f ðtÞ
st f ðtÞ
¼ e
dt ¼ FðuÞ du:
L
t
t
s
0
Letting s ! 0 þ and assuming that the integral
converges, it follows that
Z1
Z1
f ðtÞ
dt ¼ FðuÞ du:
t
0
0
For example, if f (t) ¼ sin t, then FðsÞ ¼ s2 1þ1 and so
Z1
Z1
sin t
du
¼ ½tan1 u1
dt ¼
0 ¼ :
t
u2 þ 1
2
0
at
ebt
t
:
Solution. We have
Lfeat ebt g ¼ Lfeat g Lfebt g
1
1
:
¼
sþa sþb
Therefore, proceeding as in Example 5.61, we have
Z1
1
1
uþa 1
du ¼ log
Lf f ðtÞg ¼
uþa uþb
uþb s
s
uþa
sþa
log
u!1
uþb
sþb
sþa
sþb
¼ log
:
¼ 0 log
sþb
sþa
¼ lim log
EXAMPLE 5.63
2t
:
Find the Laplace transfrom of f ðtÞ ¼ 1cos
t
0
Solution. We have
EXAMPLE 5.61
3t
:
Find the Laplace transform of f ðtÞ ¼ cos 2tcos
t
Lf1 cos 2tg ¼ Lf1g Lfcos 2tg ¼
1
s
:
s s2 þ 4
Laplace Transform
Therefore, by Theorem 5.10, we get
Z1
1 cos 2t
1
u
du
¼
2
L
t
u u þ4
s
1
1
2
logðu þ 4Þ
¼ log u 2
s
1
1
1
¼ log u2 logðu2 þ 4Þ
2
2
s
2
1
s þ4
:
¼ log
2
s2
EXAMPLE 5.64
Using Remark 5.3, evaluate the integral
Z1 t
e e3t
dt:
t
0
Z1
5.19
EXAMPLE 5.66
Find Laplace transform of f ðtÞ ¼ sint at :
Solution. We know that
Lfsin atg ¼
s2
a
:
þ a2
Therefore, by Theorem 5.10,
Z1
sin at
du
1 a
L
¼
tan
¼a
:
t
u 2 þ a2
s
s
Theorem 5.11. (Laplace Transform of Integrals). If
L{ f (t)} ¼ F(s), then
8 t
9
<Z
= FðsÞ
:
f ðuÞ du ¼
L
:
;
s
0
0
Solution. By Remark 5.3, we have
Z1 t
e e3t
dt
t
n
Proof: The function f (t) should be integrable in such
a way that
Zt
gðtÞ ¼
Z1
f ðuÞ du
0
1
1
du is of exponential order. Then g (0) ¼ 0 and g0 (t) ¼ f (t).
uþ1 uþ3
Therefore,
0
0
1
u
þ
1
Lfg 0 ðtÞg ¼ s LfgðtÞg gð0Þ ¼ s LfgðtÞg
¼ ½logðu þ 1Þ logðu þ 3Þ1
0 ¼ log
uþ3 0
and so
9
8 t
uþ1
1
log
¼ log lim
=
<Z
u!1 u þ 3
Lfg0 ðtÞg
3
¼ LfgðtÞg ¼
f
ðuÞdu
L
1 þ ð1=uÞ
1
;
:
s
log
¼ log lim
0
u!1 1 þ ð3=uÞ
3
1
Lf f ðtÞg FðsÞ
¼ log 1 log ¼ log 3:
¼
:
¼
3
s
s
¼
Lfet e3t g ds ¼
EXAMPLE 5.65
t
Find the Laplace transform of f ðtÞ ¼ 1e
t :
EXAMPLE 5.67
Rt
Find Laplace transform of
Solution. Since
Lf1 et g ¼ Lf1g Lfet g ¼
1
1
s s1 ; we have
Z1 1
1
du
Lf f ðtÞg ¼
u u1
Solution. From Example 5.66, we have
sin t
1 1
L
¼ tan
:
t
s
s
1Þ1
s
¼ ½log u logðu h
u i1
1
¼ log
¼ log
u1 s
1 ð1=sÞ
s1
:
¼ log
s
0
sin u
u du:
Therefore, by Theorem 5.11, we have
8 t
9
<Z sin u = 1
1 1
du ¼ tan
:
L
:
; s
u
s
0
(The function
Rt
0
sin u
u
du is called sine integral
function, denoted by Si(t), which is used in optics).
5.20
n
Engineering Mathematics-II
EXAMPLE 5.68
Rt
Find the Laplace transform of cos 2u du:
0
s
Solution. We know that Lfcos 2tg ¼ s2 þ4
and so
9
8 t
= 1
<Z
s
1
¼
:
L
cos 2u du ¼ : 2
; s s þ 4 s2 þ 4
:
0
EXAMPLE 5.69
Find Laplace transform of cosine integral function
Zt
cos u
CiðtÞ ¼
du; t > 0
u
1
s
: Therefore,
Solution. We know that Lfcos tg ¼ s2 þ1
by Theorem 5.10
Z1
ncos to Z1 u
1
2u
du ¼
du
L
¼
t
u2 þ 1
2 u2 þ 1
s
s
1
¼ logð1 þ s2 Þ:
2
Now Theorem 5.11 yields the Laplace transform
of cosine integral function Ci(t) as given below:
8 t
8 1
9
9
< Z cos u =
< Z cos u =
LfCiðtÞg ¼ L
du ¼ L du
:
;
:
;
u
u
1
t
9
8 1
<Z cos u =
du
¼ L
;
:
u
t
1 1
2
¼
logð1 þ s Þ
s 2
1
¼ logð1 þ s2 Þ:
2s
EXAMPLE 5.70
Find the Laplace transform of the exponential
integral defined by 1
Z u
e
du; t > 0:
EiðtÞ ¼
u
t
1
: Therefore, by
Solution. We know that Lfeu g ¼ sþ1
Theorem 5.10, we have
Z1
eu
1
¼
du ¼ logð1 þ sÞ:
L
u
uþ1
s
Now application of Theorem 5.11 yields
1
LfEiðtÞg ¼ logð1 þ sÞ:
s
EXAMPLE 5.71
Rt
Find Laplace transform of et sint t dt:
0
Solution. We know that Lfsin tg ¼ s2 1þ1 : Therefore,
Z1
sin t
1
du ¼ ½tan1 u1
¼
L
s
t
u2 þ 1
s
¼ tan1 s ¼ cot1 s:
2
Therefore, by first-shifting property, we have
L et
Hence
8 t
<Z
L
:
sin t
t
¼ cot1 ðs 1Þ:
9
= 1
sin
t
dt ¼ cot1 ðs 1Þ:
et
; s
t
0
Theorem 5.12. (Laplace Transform of a Periodic
Function). Let f be a periodic function with period T
so that f (t) ¼ f (t þ T). Then
ZT
1
est f ðtÞ dt:
Lf f ðtÞg ¼
1 esT
0
Proof: We begin with the definition of Laplace
transform and evaluate the integral using periodicity of f (t). Thus we have
Z1
Lf f ðtÞg ¼ est f ðtÞ dt
0
ZT
¼
e
st
Z2T
f ðtÞ dt þ
est f ðtÞ dt
T
0
Z3T
þ est f ðtÞ dt þ . . .
2T
ZnT
þ
ðn1ÞT
est f ðtÞ dtþ . . .
Laplace Transform
provided that the series on the right-hand side converges. This is true since the function f satisfies
the condition for the existence of its Laplace
transform. Consider the integral
ZnT
est f ðtÞ dt
0
1
¼
1 e2s
and substitute u ¼ t – (n – 1)T. Since f has period T,
we have
ZnT
ZT
st
sðn1ÞT
e f ðtÞdt¼ e
esu f ðuÞ du;
1
¼
1 e2s
Lff ðtÞg ¼ ð1 þ e
1
¼
1 esT
2sT
þe
e
st
Z
est sin t dt
1
est ðs sin t cos t
1 e2s
s2 þ 1
1
1 þ es
¼
1 e2s s2 þ 1
¼
ZT
þ ...Þ
e
st
f ðtÞdt
0
ZT
est f ðtÞ dt
0
n ¼ 1; 2; . . .
sT
Z2
0
0
Thus, we have
5.21
The function is of period 2. So, Theorem 5.12
yields
ZT
1
esT f ðtÞ dt
Lf f ðtÞg ¼
1 esT
ðn1ÞT
ðn1ÞT
n
f ðtÞdt (summing the G:P:Þ;
¼
0
1
ð1 es Þ ðs2
þ 1Þ
0
:
1
; jrj < 1.
since 1 þ r þ r2 þ r3 þ . . . ¼ 1r
EXAMPLE 5.72
Find Laplace transform of the half-wave rectified
sinusoidal
(
sin t for 2n < t < ð2n þ 1Þ
f ðtÞ ¼
0
for ð2n þ 1Þ < t < ð2n þ 2Þ;
EXAMPLE 5.73
Find the Laplace transform of full rectified sine
wave defined by the expression
(
sin t for 0 < t < f ðtÞ ¼
sin t for < t < 2
f ðtÞ ¼ f ðt þ Þ:
for n ¼ 0, 1, 2,. . ..
Solution. The graph of the half-wave rectified sinusoidal function f is shown in the Figure 5.9
Solution. The graph of the rectified sine wave is
shown in the Figure 5.10.
f (t )
f(t )
0
π
2π
Figure 5.9
3π
4π
t
0
π
2π
Figure 5.10
3π
4π
t
5.22
n
Engineering Mathematics-II
The function f has period and, therefore, application of Theorem 5.12 yields
Z
1
Lf f ðtÞg ¼
est sin t dt
1 es
EXAMPLE 5.75
Find the Laplace transform of the saw tooth wave
function, whose graph is shown in the Figure 5.12.
f(t )
0
1
1 þ es
s
s2 þ 1
1e
s
1þe
:
¼
ð1 es Þ ðs2 þ 1Þ
¼
a
EXAMPLE 5.74
Find Laplace transform of the triangular wave
function defined by
(
t
for 0 t < a
f ðtÞ ¼
2a t for 0 t < 2a;
f ð2a þ tÞ ¼ f ðtÞ:
Solution. The graph of triangular wave function is
shown in the Figure 5.11.
t
0
a
2a
3a
Figure 5.12
Solution. It is a periodic function with period a.
Therefore,
Za
Z1
1
1
st
e f ðtÞdt ¼
t est dt
Lf f ðtÞg ¼
1 eas
1 eas
0
9
8
Za st =
1 < test a
e
¼
dt
s ;
1 eas : s 0
f (t )
0
0
a
¼
a
2a
3a
t
e
s
0
1
est est
t
¼
1 eas s s2
t
0
1
1 eas
st t
4a
Figure 5.11
Using the formula for the Laplace transform of a
periodic function, we have
Z2a
1
est f ðtÞ dt
Lf f ðtÞg ¼
1 e2as
sa
est
s2
a
0
a
0
sa
¼
1
ae
e
1
2 þ 2
s
1 eas s
s
¼
1
aesa 1 esa
þ
s
s2
1 eas
¼
1
aesa
:
s2 sð1 eas Þ
0
¼
1
1 e2as
2 a
3
Z2a
Z
5.3
4 t est dt þ ð2a tÞest dt5:
0
a
Now integration by parts gives
1
1
Lf f ðsÞg ¼
ðeas 1Þ2 : 2
1 e2as
s
as
1 1 eas
1
¼
¼ 2
tanh
:
1 þ eas
s
s2
2
LIMITING THEOREMS
Theorem 5.13. (Initial Value). Let f (t) and its derivative f 0 (t) be piecewise continuous and of exponential order, then
lim f ðtÞ ¼ lim s FðsÞ;
t!0
provided the limits exist.
s!1
Laplace Transform
Proof: We know that
0
Lf f ðtÞg ¼ sFðsÞ f ð0Þ:
ð3Þ
Further, since f 0 (t) obeys the criteria for the existence of Laplace transform, we have
1
Z1
Z
st 0
e f ðtÞdt jest f 0 ðtÞj dt
0
0
Z1
est eMt dt
1
:
sþ1
lim f ðtÞ ¼ f ð0Þ ¼ 1;
t!0
s
¼ 1:
s!1
s!1 s þ 1
Thus, initial value theorem is verified in this case.
On the other hand,
lim f ðtÞ ¼ lim et ¼ 0;
lim sFðsÞ ¼ lim
lim s FðsÞ ¼ f ð0Þ ¼ lim f ðtÞ:
t!0
Theorem 5.14 (Final Value). If the limits indicated
exist, then
lim f ðtÞ ¼ lim s FðsÞ:
s!0
Proof: Using Laplace transform of the derivative, we
have
Z1
0
Lf f ðtÞg ¼ est f 0 ðtÞ dt ¼ sFðsÞ f ð0Þ ð4Þ
Z1
est f 0 ðtÞ dt ¼ lim lim
lim
ZT
s!0 T!1
s!0
0
s
¼ 0;
þ1
and so the final value theorem also holds for e–t.
EXAMPLE 5.77
Suppose Laplace transform of a function f is
given by
18
FðsÞ ¼ 2
:
sðs þ 36Þ
Find f (0) and f 0 (0).
Solution. By initial value theorem, we have
f ð0Þ ¼ lim f ðtÞ ¼ lim s FðsÞ
s!1
s!0
18 s
¼ 0;
¼ lim 2
s!1 sðs þ 36Þ
and
0
Then
s!0 s
s!0
s!1
t!1
t!1
lim sFðsÞ ¼ lim
0
1
! 0 as s ! 1:
M s
Hence (3) implies s F(s) – f (0) ! 0 as s ! 1, and so
5.23
Solution. We observe that
Lfet g ¼ FðsÞ ¼
t!1
¼
n
f 0 ð0Þ ¼ lim ½s2 FðsÞ s f ð0Þ
s!1
est f 0 ðtÞ dt
18s2
0 ¼ 0:
þ 36Þ
¼ lim
s!1 sðs2
0
¼ lim lim fesT f ðT Þ f ð0Þg
s!0 T !1
¼ lim f ðT Þ f ð0Þ
T!1
¼ lim f ðtÞ f ð0Þ:
t!1
Hence relation (4) yields
EXAMPLE 5.78
1
Let FðsÞ ¼ s1
: Can we find f (1) using final value
theorem?
Solution. In this case, the final value theorem does
not apply because
lim f ðtÞ ¼ lim et ! 1 as t ! 1:
t!1
lim f ðtÞ f ð0Þ ¼ lim sFðsÞ f ð0Þ
t!1
t!1
s!0
5.4
and so
lim f ðtÞ ¼ lim sFðsÞ:
t!1
s!0
EXAMPLE 5.76
Verify the validity of limiting theorems by considering f (t) ¼ et.
MISCELLANEOUS EXAMPLES
EXAMPLE 5.79
Find the Laplace Transforms of
(i) e4t sin 2t cos t;
(ii) sinh t cos2 t and
(iii) et cos2 t:
5.24
n
Engineering Mathematics-II
Solution. (i) We have
Therefore, by shifting property,
L et cos2 t ¼
1
sin 2t cos t ¼ ½sin 3t þ sin t:
2
Therefore
¼
1
1
Lfsin 2t cos tg ¼ Lfsin 3tg þ Lfsin tg
2
2
1
3
1
¼
þ
:
2 s2 þ 9 s2 þ 1
Now using shifting property, we get
L e4t sin 2t cos t
"
#
1
3
1
¼
þ
2 ðs 4Þ2 þ 9 ðs 4Þ2 þ 1
¼
1
3
1
þ
:
2 s2 8s þ 25 s2 8s þ 17
(ii) Since cos t ¼
2
1þcos 2 t
.
2
1
s2 2s þ 3
¼
2 ðs 1Þðs2 2s þ 5Þ
s2 þ 2s þ 3
:
ðs þ 1Þðs2 þ 2s þ 5Þ
(iii) As in part (ii),
s2 þ 2
; s > 0:
L cos2 t ¼ 2
sðs þ 4Þ
s2 þ 2s þ 3
:
ðs þ 1Þðs2 þ 2s þ 5Þ
Solution. (i) Using linearty and shifting properties,
we have
L e3t ð2 cos 5t 3 sin 5tÞ
¼ 2L e3t cos 5t 3L e3t sin 5t
¼2
1
1
L cos2 t ¼ Lf1g þ Lfcos 2tg
2
2
1 1
s
s2 þ 2
¼ 2
; s > 0:
þ 2
¼
2 s s þ4
sðs þ 4Þ
et et
L sinh t cos2 t ¼ L
cos2 t
2
1 1 ¼ L et cos2 t L et cos2 t
2(
2
)
1
ðs 1Þ2 þ 2
¼
2 ðs 1Þ½ðs 1Þ2 þ 4
(
)
1
ðs þ 1Þ2 þ 2
2 ðs þ 1Þ½ðs þ 1Þ2 þ 4
ðs þ 1Þ½ðs þ 1Þ2 þ 4
EXAMPLE 5.80
Find the Laplace transforms of the following:
(i) e3t ð2 cos 5t 3 sin 5tÞ and
(ii) 1t ðcos at cos btÞ:
Therefore
Therefore, by shifting property,
ðs þ 1Þ2 þ 2
¼
¼
sþ3
2
ðs þ 3Þ þ 25
3
5
ðs þ 3Þ2 þ 25
2s 9
ðs þ 3Þ2 þ 25
2s 9
:
s2 þ 6s þ 34
(ii) As in Example 5.64, we have
Z1
cos at cos bt
dt
t
0
Z1
¼
Lfcos at cos btgds
0
Z1
¼
s
s
ds
s 2 þ a2 s 2 þ b 2
0
¼
1
1
logðs2 þ a2 Þ logðs2 þ b2 Þ
2
2
2 1
1
s þa
log 2
s þ b2 0
2
" (
)#
2
1 þ as2
1
1
a2
¼
log lim
log
2
s!1 1 þ b2
b2
2
2
s
¼
2
1
a2 1
b2
¼ log 2 ¼ log 2 :
b
a
2
2
1
0
n
Laplace Transform
EXAMPLE 5.81
Find the Laplace transform of cosh at sin at.
EXAMPLE 5.83
Evaluate:
5.25
L te2t sin 4t
Solution. Let
gðtÞ ¼ cosh at sin at:
Solution. We know that
Then
1
:
þ 16
Now, by shifting property,
1
Lfe2t sin 4tg ¼
:
ðs þ 2Þ2 þ 16
Lfsin 4tg ¼
LfgðtÞg ¼ L
eat þ eat
sin at
2
1
¼ Lfeat sin at þ eat sin atg
2
1
1
¼ Lfeat sin atg þ Lfeat sin atg:
2
2
Now
Lfsin atg ¼
s2
s2
Further,
"
#
d
1
Lfe2t sin 4tg ¼ ds ðs þ 2Þ2 þ 16
a
:
þ a2
¼
2s þ 4
ðs2 þ 4s þ 20Þ2
:
Therefore, by shifting property,
Lfeat sin atg ¼
Lfeat sin atg ¼
a
2
ðs aÞ þ
a
a2
ðs þ aÞ2 þ a2
;
:
EXAMPLE 5.84
9
8
< 1 if 0 < t 1 =
Express f ðtÞ ¼
t if 1 < t 2 in terms of
;
: 2
t if
t>2
unit step function and hence find Lf ðtÞ.
Hence
"
#
1
a
a
þ
LfgðtÞg ¼
2 ðs aÞ2 þ a2 ðs þ aÞ2 þ a2
EXAMPLE 5.82
Find the Laplace transform of f ðtÞ ¼ t2 sin2 at.
Solution. Let f ðtÞ ¼ t2 sin2 at: We know that
2a2
i:
L sin2 at ¼ h
s s2 þ ð2aÞ2
Therefore
d2
2a2
L t2 sin2 at ¼ ð1Þ2 2
ds sðs2 þ 4a2 Þ
2
d
1
¼ 2a2 2
2
ds sðs þ 4a2 Þ
"
#
4
2 2
4
2 3s þ 6a s 8a
¼ 8a
:
s2 ðs2 þ 4a2 Þ3
Solution. In terms of Heavyside s unit step function,
we have
f ðtÞ ¼ HðtÞ þ tHðt 1Þ þ t2 Hðt 2Þ:
Therefore
Lff ðtÞg ¼ LfHðtÞg þ LftHðt 1Þg þ L t2 Hðt 2Þ
1 d es
d e2s
þ ð1Þ2
¼ s ds s
ds s
1 es es e2s 2e2s
2 2 ¼ s
s
s
s
s
1
1 s
s
2s
¼ ½1 e 2e 2 ½e þ e2s :
s
s
EXAMPLE 5.85
Find the Laplace Transform of e3t ½t cos 2t and
1e2t
t .
Solution. We know that
FðsÞ ¼ Lft cos 2tg ¼
s2 4
ðs2 þ 4Þ2
:
5.26
n
Engineering Mathematics-II
Therefore, by shifting property,
Therefore
Lfe ðt cos2tÞg ¼ Fðs 3Þ
3t
¼
ðs 3Þ2 4
ððs 3Þ2 þ 4Þ
2
¼
s2 6s þ 5
ðs2 6s þ 13Þ2
Lfe
:
Also, As in Example 5.65,
Lf1 e2t g ¼ Lf1g Lfe2t g ¼
Therefore,
Z1 Lff ðtÞg ¼
1
1
:
s sþ2
1
1
du
u uþ2
s
¼ ½log u logðu þ 2Þ1
s "
u 1
1
¼ log
¼ log
uþ2 s
1 þ 2s
sþ2
¼ log
:
s
EXAMPLE 5.86
Evaluate:
L
"
#
1
1
sþa
sin tg ¼
:
2 s þ a ðs þ aÞ2 þ 4
2
Further,
Z1 at 2
Z1
e sin t
dt ¼ Lfeat sin2 tgds
t
0
0
2 1
3
Z
Z1
14
ds
sþa
¼
ds5
2
sþa
ðs þ aÞ2 þ 4
0
20
31
16
sþa
7
¼ 4log qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi5
2
2
ðs þ aÞ þ 4
#
1 cos at
t
0
1
1
a
¼ 0 log pffiffiffiffiffiffiffiffiffiffiffiffiffi
2
2
a2 þ 4
pffiffiffiffiffiffiffiffiffiffiffiffiffi
1
a2 þ 4
¼ log
:
2
a
EXERCISES
Solution. See Example 5.63. We have
1
s
Lf1 cos atg ¼ Lf1g Lfcos a tg ¼ 2
:
s s þ a2
Therefore
Z1
1 cos at
1
u
du
¼
L
t
u u2 þ a 2
s
1
1
logðu2 þ a2 Þ
¼ log u 2
s
1
1
1
2
2
2
¼ log u logðu þ a Þ
2
2
s
1
s 2 þ a2
¼ log
:
s2
2
EXAMPLE 5.87
R1
Using Laplace transform evaluate
0
Solution. We have
1
Lfsin2 tg ¼ L ð1 cos 2tÞ
2
at
¼
eat sin2 t
dt.
t
1 1
s
:
2
2 s s þ4
1. Find the Laplace transforms of
6
Ans. s42 þ s4
(a) 4t þ 6e4t
5
Ans. s2 þ8sþ41
(b) e–4t sin 5t
(c)1t ðsin atat cos atÞ Ans. tan1 as s2 as
þa2
s
(d) 2at sin at
Ans. ðs2 þa
2
2Þ
(e)
tx1 eat
ðxÞ
( f ) f ðtÞ ¼
Ans.
tþ1
3
1
ðsaÞx
;x>0
for 0 t 2
for t > 2
Ans. 1s þ s12 ðe2s 1Þ
(g) f (t) ¼ t H(t – a)
Ans.
ð1þasÞ eas
s2
Rt
(h) Null function defined by nðtÞ dt ¼ 0 for
0
all t.
Ans. 0
2. Show that the Laplace transforms of the following functions do not exist:
2
Hint: Not of exponential order
(a) et
1/t
Hint: Not defined at t ¼ 0
(b) e
(c) f ðtÞ ¼
1 for even t
0 for odd t
n
Laplace Transform
Hint: has infinite number of finite jumps and so
condition of piecewise continuity is not satisfied
(d) t–n, n is positive integer.
3. Find L sinht vt
Ans. 12 log sþv
sv ; s > jvj
¼
est 2
est 3
est 4
þ2
þ3
þ...
s 1
s 2
s 3
es
ð1 þ es þ e2s þ . . .Þ
s
es
:
¼
sðes 1Þ
¼
4. Find the Laplace transform of a function whose
graph is shown in the Figure 5.13.
f(t )
6. Find the Laplace transforms of
Ans.
(a) t e–2t sin 2t
(b) t2 cos at
(c) t sin 3t cos 2t.
1
t
0
1
2
Figure 5.13
Hint: The function is defined by
t
for 0 t < 1
f ðtÞ ¼
2 t for 1 t < 2:
R2
R1
Therefore,
Lf f ðtÞg ¼ test dt þ ð2 tÞ
0
1
est dt ¼ s12 ð1 es Þ2 .
5. Find the Laplace transform of step function f
defined by f (t) ¼ n, n t < n þ 1, n ¼ 0, 1, 2, . . .
Hint: The graph of step function is shown in the
Figure 5.14.
Therefore,
Lfsin 3t cos 2tg ¼
¼
n
f ðtÞ
t
o
¼
R1
t
1
2
3
Z1
Lf f ðtÞg ¼ est f ðtÞ dt
0
e
1
þ 25Þ
st
Z3
dt þ 2
e
2
st
Z4
dt þ 3
3
2
þ
s
ðs2
þ 1Þ2
:
Ans. cot1 ðs þ 1Þ.
R1 et sin2 t
t
s2 36
sa
dt.
Hint: I ¼
Figure 5.14
¼
5s
ðs2
Ans. log
0
Z2
d
ð3s2 þ 15Þ
2
ds ðs þ 1Þ ðs2 þ 25Þ
FðsÞ ds.
s
eat cos 6t
t
8. Evaluate I ¼
0
2s3 6a2 s
ðs2 þa2 Þ3
7. Find Laplace transforms of
t
(a) e tsin t Hint: First find L{e–tsin t} and then use
(b)
1
Ans.
3s2 þ 15
; s > 0 and so
ðs2 þ 1Þ ðs2 þ 25Þ
Lfs sin 3t cos 2tÞ ¼ L
2
4ðs2Þ
ðs2 4sþ8Þ2
1
Hint : sin 3t cos 2t ¼ ð2 sin 3t cos 2tÞ
2
1
1
1
¼ ðsinð3t þ 2tÞþ sinð3t 2tÞ ¼ sin 5t þ sin t:
2
2
2
f (t )
3
5.27
R1
Lf f ðtÞg du; where
0
f (t) ¼ e–t sin2 t. But Lfet sin2 tg ¼ ðsþ1Þðs22 þ2sþ5Þ :
Therefore,
R1
Ans. 12 log 5
I ¼ ðsþ1Þ ðs22 þ2sþ5Þ ds
0
9. Show that Lftf 0 ðtÞg ¼ s dsd FðsÞ þ FðsÞ :
10. Show that L 1t ðsin at at cos atÞ ¼ tan1 as as
s2 þa2 :
est dt þ . . .11. Show that Lft2 f 00 ðtÞg ¼ s2 d 22 FðsÞ þ 4s d FðsÞþ
ds
ds
þ 2FðsÞ:
5.28
n
Engineering Mathematics-II
12. Use initial value theorem to find f (0) and f 0 (0)
s
for the function f for which FðsÞ ¼ s2 5sþ12
16. Solve Example 5.27 using periodicity
R2a
Hint : Lf f ðtÞg ¼ 1e12as est f ðtÞ dt ¼ 1e12as
Ans. f (0) ¼ 1, f 0 (0) ¼ 5
13. Find the Laplace transform of the square wave
function with graph shown in the Figure 5.15.
0
f(t )
t
a
2a
3a
4a
5a
Figure 5.15
Hint: The function is of period 2a. Therefore,
Z2a
1
eas
Lf f ðtÞg ¼
est dt ¼
2as
1e
sð1 þ eas Þ
a
14. Express the function f (t) in Exercise 13 in terms
of Heaviside’s unit step function and then find
its Laplace transform.
Hint: f (t) ¼ H(t – a) – H(t – 2a) þ H(t þ 3a) –
H(t – 4a) þ . . .
15. Find the Laplace transform the of square wave
function with graph given in the Figure 5.16.
f (t )
1
0
−1
est dt þ
R2a
0
est dt
which
on
using
a
Examples 5.13 and 5.15 yields the transform as
1
as
tanh :
s
2
1
0
Ra
a
2a
3a
4a
t
Figure 5.16
Hint: Here the function is of period 2a.
Therefore,
Za
1
Lf f ðtÞg ¼
est dt
1 e2as
0
1
est a
1
¼
¼
2as
s 0 sð1 þ eas Þ:
1e
17. Find Laplace transform of the half-wave rectified sine function f defined by
(
ð2nþ1Þ
sin vt for 2n
v <t <
v
f ðtÞ ¼
ð2nþ2Þ
0
for ð2nþ1Þ
<
t
<
v
v
Hint: The function is periodic with period
Therefore,
2=v
Z
1
est sin vt dt
Lf f ðtÞg ¼
1 e2s=v
2
v.
0
Z=v
1
est sin vt dt
1 e2s=v
0
v
ð1 þ es=v Þ:
¼ 2
s þ v2
¼
18. Establish relation between Laplace transform
and Fourier transform of a function.
Hint: Let be a function defined by
ext f ðtÞ for t > 0
ðt Þ ¼
0
for t < 0:
Then the Fourier transform of is
Z1
F ð yÞ ¼ F fðtÞg ¼ eiyt :ext f ðtÞdt
0
Z1
¼ eðxþiyÞt f ðtÞdt
0
Z1
¼ est f ðtÞdt ¼ Lf f ðtÞg:
0
Thus Laplace transform
of f (t) is equal to the
Fourier transform of extf (t).
6
Inverse Laplace Transform
Like the operations of addition, multiplication, and differentiation, the Laplace transform has also its inverse.
During the process of solving physical problems like
differential equations, it is necessary to invoke the
inverse transform of the Laplace transform. Thus
given a Laplace transform F(s) of a function f, we
would like to know what f is. Hence, we are concerned
with the solution of the integral equation,
Z1
est f ðtÞ dt ¼ FðsÞ:
(a) Let f (t) ¼ sin vt, t
v
Lff ðtÞg ¼ s2 þv
2 . Thus
L1
gðtÞ ¼
sin vt
1
Definition 6.1. Let f have Laplace transform F(s), that
is, L{ f(t)} ¼ F(s), then f(t) is called an inverse
Laplace transform of F(s) and we write
L1 fFðsÞg ¼ f ðtÞ; t
0:
1
The transformation L is called inverse Laplace
operator and it maps the Laplace transform of a
function back to the original function.
We know that Laplace transform F(s) of a
function f(t) is uniquely determined due to the
properties of integrals. However, this is not true for
the inverse transform. For example, if f(t) and g(t)
are two functions that are identical except for a
finite number of points, they have the same transform F(s) since their integrals are identified.
Therefore, either f(t) or g(t) is the inverse transform
of F(s). Thus inverse transform of a given function
F(s) is uniquely determined only upto an additive
Rt
null function [a function n(t) for which nðuÞ du ¼
0
0 for all t]. The following examples show that L1
{F(s)} can be more than one function.
for t > 0
for t ¼ 0:
Then
¼
v
;
s2 þ v2
v
þ v2
¼ gðtÞ:
LfgðtÞg
DEFINITION AND EXAMPLES OF INVERSE
LAPLACE TRANSFORM
¼ sin vt:
Now let
0
6.1
v
s2 þ v 2
0. Then
and so
L1
s2
v
Hence there are two inverse transforms of s2 þv
2.
3t
(b) Let f (t) ¼ e and
gðtÞ ¼
0
e3t
for t ¼ 1
otherwise:
Then both f(t) and g(t) have same Laplace transform
1
1
sþ3. Thus sþ3 has two inverse Laplace transforms f(t)
and g(t).
But the following theorem shows that the
Laplace transform is one-one mapping.
Theorem 6.1. (Lerch’s Theorem). Distinct continuous
functions on [0, 1) have distinct Laplace
transforms.
Thus, if we restrict ourselves to continuous
functions on [0, 1), then the inverse transform
L1{F(s)} ¼ f(t) is uniquely defined. Since many of
the functions, we generally deal with, are solutions
to the differential equations and hence continuous,
the assumption of the theorem is satisfied.
6.2
n
Engineering Mathematics-II
EXAMPLE 6.1
Find inverse Laplace transform of
1 esa
s
n!
sa ; s ; s2 a2 and snþ1 .
a
s
s2 þa2 ; s2 þa2
;
Solution. We know that
a
s
Lfsin atg ¼ 2
; Lfcos atg ¼ 2
;
s þ a2
s þ a2
1
esa
;
; Lfhðt aÞg ¼
Lfeat g ¼
s
sa
s
; and
Lfcosh atg ¼ 2
s a2
n!
Lftn g ¼ nþ1 ; n being non-negative integer:
s
Therefore,
L1
a
¼ sin at;
s2 þa2
L1
s
¼ cos at;
s2 þa2
L1
Theorem 6.2. (Linearity Property). If F1(s) and F2(s)
are Laplace transforms of f1(t) and f2(t), respectively, and a1 and a2 are arbitrary constants, then
L1 fa1 F1 ðsÞ þ a2 F2 ðsÞg
¼ a1 L1 fF1 ðsÞg þ a2 L1 fF2 ðsÞg
¼ a1 f1 ðtÞ þ a2 f2 ðtÞ:
Proof: Since
Lfa1 f1 ðtÞ þ a2 f2 ðtÞg ¼ a1 Lff1 ðtÞg þ a2 Lff2 ðtÞg
¼ a1 F1 ðsÞ þ a2 F2 ðsÞ;
we have
L1 fa1 F1 ðsÞ þ a2 F2 ðsÞg
¼ a1 f1 ðtÞ þ a2 f2 ðtÞ
¼ a1 L1 fF1 ðsÞg þ a2 L1 fF2 ðsÞg:
EXAMPLE 6.3
Find the inverse Laplace transform of
1
¼ eat ;
sa
1
4
s
:
þ
þ
2s 3ðs aÞ s2 þ 16
esa
¼ Hðt aÞ; Heavyside’s unit step function. Solution. By linearity of inverse Laplace transform,
s
we have
n s o
1
4
s
¼ cosh at;
L1 2
L1
þ
þ 2
s a2
L1
2s
and
L1
n!
snþ1
Solution. Since L
n
1
t1=2
o
¼
¼ ð1=2Þ
s1=2
1
L1 pffiffi
s
pffiffi
s , it follows that
1
¼ pffiffiffiffiffi :
t
þ L1
PROPERTIES OF INVERSE LAPLACE TRANSFORM
The operational properties used in finding the
Laplace transform of a function are also used in constructing the inverse transform. We, thus, have the
following properties of inverse transform.
s2
s
þ 16
¼
1 4 at
þ e þ cos 4t:
2 3
EXAMPLE 6.4
Find inverse Laplace transform of
5
s
3
þ
þ 2
:
s3 s þ4 s7
Solution. By linearity, we have
L1
6.2
s þ 16
1
1
4
1
¼ L1
þ L1
2
s
3
sa
¼ tn ; n being non-negative integer,
EXAMPLE 6.2
Find L1 p1ffis .
3ðs aÞ
5
s
3
þ
þ 2
s3 s þ4 s7
¼ 5L1
1
s
þ L1 2
s3
s þ4
þ 3L1
1
s7
¼ 5e3t þ cos 2t þ 3 e7t :
EXAMPLE 6.5
Find inverse Laplace transform of
3
4
1
þ
:
þ
2s 3ðs 1Þ 6ðs þ 2Þ
Solution. By linearity of inverse Laplace transform,
we have
3
4
1
þ
L1 þ
2s 3ðs 1Þ 6ðs þ 2Þ
EXAMPLE 6.7
s5
.
Find inverse Laplace transform of s2 þ6sþ13
Solution. Since
s2
s5
ðs þ 3Þ 8
¼
;
þ 6s þ 13 ðs þ 3Þ2 þ 4
we have
1
L
3
1
4
1
1
1
þ L1
þ L1
¼ L1
2
s
3
s1
6
sþ2
s5
2
s þ 6s þ 13
ðs þ 3Þ2 þ 4
EXAMPLE 6.8
2s3
.
Find inverse Laplace transform of s2 þ4sþ13
pffiffi
1
1
a
1
að s aÞ
pffiffi
¼ pffiffi pffiffi pffiffi
¼ pffiffi pffiffi
;
sþa
s
sð s þ aÞ
s
sðs a2 Þ
L
ðs þ 3Þ2 þ 4
(
)
2
1
¼ e3t ðcos 2t 4 sin 2tÞ:
Solution. Since
1
pffiffi
sþa
¼L
)
sþ3
¼ e3t cos 2t 4e3t sin 2t
EXAMPLE 6.6
n
o
Find L1 pffis1þa .
1
(
1
4 L
3 4
1
¼ þ et þ e2t :
2 3
6
therefore,
6.3
n
Inverse Laplace Transform
pffiffi
1
sa
1
pffiffi aL
pffiffi
¼L
s
sðs a2 Þ
1
1
¼ L1 pffiffi aL1
s
a2
s
1
þ a2 L1 pffiffi
sðs a2 Þ
pffi
1
2
2
¼ pffiffiffiffiffi aea t þ aea t erf ða tÞ
t
pffi
1
2
¼ pffiffiffiffiffi aea t ð1 erf ða tÞ
t
pffi
1
2
¼ pffiffiffiffiffi aea t erfc ða tÞ
t
1
Theorem 6.3. (First Shifting Property). If F(s) is
Laplace transform of f(t), then
L1 fFðs aÞg ¼ eat f ðtÞ:
Proof: We know that
Lfeat f ðtÞg
¼ Fðs aÞ:
Therefore,
L1 fFðs aÞg ¼ eat f ðtÞ:
Solution. Since
s2
2s 3
2s þ 4 7
2ðs þ 2Þ 7
¼
¼
;
2
þ 4s þ 3 ðs þ 2Þ þ 9 ðs þ 2Þ2 þ 9
we have
1
L
2s 3
2
s þ 4s þ 13
(
1
¼L
7 L
2ðs þ 2Þ
)
ðs þ 2Þ2 þ 9
(
1
1
)
ðs þ 2Þ2 þ 9
7
¼ 2e2t cos 3t e2t sin 3t:
3
EXAMPLE 6.9
s
Find inverse Laplace transform of ðsþ1Þ
2.
Solution. We note that
s
ðs þ 1Þ
2
¼
sþ11
ðs þ 1Þ
2
¼
1
1
:
s þ 1 ðs þ 1Þ2
Therefore,
(
)
(
)
s
1
1
1
1
1
L
L
¼ L
sþ1
ðs þ 1Þ2
ðs þ 1Þ2
¼ et :1 et :t ¼ et ð1 tÞ:
6.4
n
Engineering Mathematics-II
EXAMPLE 6.10
Find inverse Laplace transform of
1
sþ4
sþ2
þ
:
s2 þ 4s þ 13 s2 þ 8s þ 97 s2 4s þ 29
Solution. We have
FðsÞ ¼
¼
Therefore, by linearity of inverse Laplace transform
and shifting property, we get (
)
s
þ
2
s
2
¼ L1
L1 2
s 4s þ 13
ðs 2Þ2 þ 9
(
)
4 1
3
þ L
3
ðs 2Þ2 þ 9
1
sþ4
sþ2
þ
s2 þ 4s þ 13 s2 þ 8s þ 97 s2 4s þ 29
4
¼ e2t cos 3t þ e2t sin 3t
3
4
¼ e2t ðcos 3t þ sin 3tÞ:
3
1
ðs þ 2Þ2 þ 9
sþ4
ðs þ 4Þ2 þ 81
þ
s2þ4
ðs 2Þ2 þ 25
Theorem 6.4. (Second Shifting Property). If
L1 {F(s)} ¼ f(t), then L1 {esa F(s)} ¼ g(t), where
:
f ðt aÞ
0
gðtÞ ¼
Therefore,
(
)
1
3
L1 fFðsÞg ¼ L1
3
ðs þ 2Þ2 þ 9
(
)
sþ4
1
L
ðs þ 4Þ2 þ 81
(
)
s2
1
þL
ðs 2Þ2 þ 25
(
)
4 1
5
þ L
5
ðs 2Þ2 þ 25
Proof: Since L{g(t)} ¼ esa F(s), it follows that
L1 fesa FðsÞg ¼ gðtÞ:
Second Proof. By definition of Laplace transform, we
have
Z1
FðsÞ ¼
est f ðtÞ dt:
0
Therefore,
e
sa
Solution. We have
s2
¼
ðs 2Þ
2
ðs 2Þ þ 9
s2
þ
4
2
ðs 2Þ þ 9
4
3
þ :
3
ðs 2Þ þ 9
ðs 2Þ2 þ 9
2
e
sa st
e
Z1
f ðtÞ dt ¼
0
Z1
¼
0
esu f ðu aÞ du;
a
Za
¼
esðtþaÞ f ðtÞ dt
e
su
Z1
ð0Þ du þ
t þ a ¼ u:
esu f ðu aÞ du
a
0
¼ LfgðtÞg:
Hence
L1 fesa FðsÞg ¼ gðtÞ:
EXAMPLE 6.12
s=2
Find inverse Laplace transform of es2 þ1 .
sþ2
ðs 2Þ þ 4
¼
4s þ 13 ðs 2Þ2 þ 9
¼
Z1
FðsÞ ¼
1
¼ e2t sin 3t e4t cos 9t
3
4
þ e2t cos 5t þ e2t sin 5t:
5
EXAMPLE 6.11
sþ2
.
Find the inverse Laplace transform of s2 4sþ13
for t > a
for t < a:
Solution. We have
es=2
1
2
¼ es=2 2
s þ1
s þ1
¼ es=2 FðsÞ; where FðsÞ ¼ s2
1
:
þ1
Inverse Laplace Transform
But
L1 fFðsÞg ¼ L1 1
s2 þ 1
¼ L1
1
:
s2 þ 1
Therefore, by second-shifting property,
L
1
es=2
2
s þ1
¼ gðtÞ;
where
sinðt =2Þ for t > =2
gðtÞ ¼
0
for t < =2
h i
¼ sin t H t
2
h 2 i
¼ cos t H t ;
2
where H(t) denotes Heavyside’s unit step function.
EXAMPLE 6.13
sa
Find inverse Laplace transform of sve
2 þv2 .
v
s2 þ v 2
Solution. We have
1
¼ 1 and
L1
s
L1
s
s2 þ 1
¼ cos t:
Therefore, using linearity property and secondshifting property, we have
A
s
esa
L1
s s2 þ 1
A
s
¼ A gðtÞ;
L esa 2
¼ AL1
s
s þ1
where
(
gðtÞ ¼
cosðt aÞ
for t > a
0
for t < a:
¼ Hðt aÞ cosðt aÞ:
Hence
A
s
esa
2
s s þ1
L1
¼ sin vt:
¼ A H ðt aÞ cosðt aÞ:
Therefore, by second-shifting property,
v
¼ gðtÞ;
L1 esa 2
s þ v2
EXAMPLE 6.16
e7s
Find the inverse Laplace transform of ðs3Þ
3.
where
Solution. Since
sin vðt aÞ for t > a
0
for t < a:
gðtÞ ¼
L
1
1
ðs 3Þ3
!
1
¼ t2 e3t ;
2
EXAMPLE 6.14
s
Find inverse transform of 2es3 , Re(s) > 0.
by second-shifting property, we have
(
)
1
1
7s
e
L
¼ gðtÞ;
ðs 3Þ3
Solution. We have
where
Since L1
property,
1
s3
2es
1
¼ 2es : 3 :
s3
s
¼ t2 , therefore, by second-shifting
L1
s
2e
s3
¼ gðtÞ;
where
gðtÞ ¼
2ðt 1Þ2
0
for t 1
for 0 t < 1:
6.5
EXAMPLE 6.15
n
o
s
esa , where A is a constant.
Find L1 As s2 þ1
Solution. We have
L1
n
(
gðtÞ ¼
1
2 ðt
7Þ2 e3ðt7Þ
0
for t > 7
for 0 t 7
1
¼ Hðt 7Þðt 7Þ2 e3ðt7Þ :
2
Theorem 6.5. (Change of Scale Property). If
L1 {F(s)} ¼ f(t), then
1 t
L1 fFðasÞg ¼ f
:
a
a
6.6
n
Engineering Mathematics-II
Proof: By the definition of Laplace transform,
Z1
FðsÞ ¼
est f ðtÞ dt:
Z1
FðasÞ ¼
0
1
¼
a
east f ðtÞ dt
Z1
esu f
u
a
0
n
o
1
t
:
¼ L f
a
a
Hence
du; u ¼ at
1 t
L fFðasÞg ¼ f
:
a
a
Remark 6.1. It follows from
6.5 that if
Theorem
L1{F(s)} ¼ f(t), then L1 F as ¼ af (at) for a > 0.
EXAMPLE 6.17
Find the inverse transform of ðs=2sÞ2 þ4.
L
we have
(
L1
s
ðs=2Þ þ 4
(
1
1
L1
sþa
sþb
¼ t f ðtÞ;
¼ 2L1
s=2
ðs=2Þ2 þ 4
Proof: Since
Lftn f ðtÞg ¼ ð1Þn
dn
FðsÞ;
dsn
n
o
L1 F ðnÞ ðsÞ ¼ ð1Þn tn f ðtÞ
eat ebt ¼ t f ðtÞ:
)
Theorem 6.6. (Inverse Laplace Transform of
Derivatives). If L1{F(s)} ¼ f(t), then
n
o
L1 F ðnÞ ðsÞ ¼ ð1Þn tn f ðtÞ:
EXAMPLE 6.18
n
o
2
.
Find L1 ðs1s
2 þ1Þ2
Solution. We note that
d
sþa d
log
¼ ½logðs þ aÞ logðs þ bÞ
ds
s þ b ds
1
1
:
¼
sþa sþb
Therefore, the use of Theorem 6.6 yields
1
1
L1
¼ t f ðtÞ
sþa sþb
that is,
¼ 2:2 cosð2ð2tÞÞ ¼ 4 cos 4t:
we have
d
s
1 s2
¼
:
ds s2 þ 1
ðs2 þ 1Þ2
Therefore, by Theorem
6.6,
(
) we have
2
1s
L1
¼ t cos t:
ðs2 þ 1Þ2
L1
¼ cos 2t;
)
2
¼ cos t:
and so
Solution. Since
s
s2 þ 4
s
s2 þ 1
EXAMPLE 6.19
n
o
Find L1 log sþa
sþb .
1
1
L1
Further
0
Therefore,
Solution. We know that
Hence
1
f ðtÞ ¼ ðebt eat Þ:
t
EXAMPLE 6.20
n
o
2
2
Find L1 log ss2 þa
þb2 .
Solution. Since
d
s 2 þ a2
d
log 2
¼ ½logðs2 þ a2 Þ logðs2 þ b2 Þ
s þ b2 ds
ds
2s
2s
2
;
¼ 2
2
s þa
s þ b2
Theorem 6.6 yields
2s
2s
¼ t f ðtÞ;
2
L1 2
2
s þa
s þ b2
or
2 cos at 2 cos bt ¼ t f ðtÞ;
Inverse Laplace Transform
or
2
f ðtÞ ¼ ðcos bt cos atÞ:
t
EXAMPLE 6.21
Find L1 log 1þs
s .
or
f ðtÞ ¼
1
L
and so
L1 1
s2 þ 1
¼ tf ðtÞ;
Hence
f ðtÞ ¼
sin t
:
t
EXAMPLE 6.24
n
o
s2 þ1
Find L1 log ðs1Þ
2 .
Solution. Since
d
s2 þ 1
log
ds
ðs 1Þ2
!
1 et
:
t
¼
d
½logðs2 þ 1Þ 2 logðs 1Þ
ds
¼
2s
2
;
s2 þ 1 s 1
we have
EXAMPLE 6.22
Find L1 log 1 þ s12 .
we have
it follows that
1ðsin tÞ ¼ t f ðtÞ:
e1 1 ¼ tf ðtÞ
Solution. Since
d
s2 þ 1
log 2
ds
s
6.7
that is,
Solution. Since
d
1þs d
log
¼ ½logð1 þ sÞ log s
ds
s
ds
1
1
;
¼
sþ1 s
Therefore,
1
1
L1
¼ tf ðtÞ
sþ1 s
or
n
L1
s2
2s
2
þ1 s1
¼ tf ðtÞ;
which yields
d
¼ ½logðs2 þ 1Þ log s2 ds
2s
2s
s
1
¼ 2
¼2 2
;
s þ 1 s2
s þ1 s
s
1
2 2
s þ1 s
¼ tf ðtÞ
2 cos t 2 ¼ t f ðtÞ
or
f ðtÞ ¼
2ð1 cos tÞ
:
t
EXAMPLE 6.23
Find L1 tan1 1s ’s > 0,
Solution.
Since d
1
1
1
tan1
¼
ds
s
s2
1 þ ð1=sÞ2
1
1
1
¼ 2
;
¼
1 þ ð1=s2 Þ s2
s þ1
2½cos t et ¼ t f ðtÞ
or
f ðtÞ ¼
2ðet cos tÞ
:
t
EXAMPLE 6.25
n
o
sþ2
Find L1 ðs2 þ4sþ5Þ
2 .
Solution. We have
sþ2
ðs2 þ 4s þ 5Þ2
Therefore,
(
L1
sþ2
ðs2 þ 4s þ 5Þ2
¼
sþ2
ððs þ 2Þ2 þ 1Þ2
)
(
¼ L1
sþ2
)
ððs þ 2Þ2 þ 1Þ2
(
)
s
2t 1
¼ e :L
:
ðs2 þ 1Þ2
n
o
s
We now find L1 ðs2 þ1Þ
2 . We note that
d
1
2s
¼
:
2
2
ds ðs þ 1Þ
ðs þ 1Þ2
6.8
n
Engineering Mathematics-II
Therefore,
by Theorem
6.6
(
)
2s
1
¼ t L1 2
¼ t sin t
L1 s þ1
ðs2 þ 1Þ2
and so
(
)
s
1
L1
¼ t sin t:
2
2
2
ðs þ 1Þ
Hence (
)
sþ2
1
1
¼ t e2t sin t:
L
2
2
2
ðs þ 4s þ 5Þ
EXAMPLE 6.26
n
o
sþ3
Find L1 ðs2 þ6sþ13Þ
.
2
Solution. We have
sþ3
f ðtÞ
L
t
we have
L1
8 1
<Z
:
Z1
¼
FðuÞ du;
s
FðuÞ du
s
9
=
;
¼
f ðtÞ
:
t
EXAMPLE 6.27
Find
8 1
9
=
<Z 1
1
du :
L1
:
;
u1 uþ1
s
sþ3
¼
:
ðs2 þ 6s þ 13Þ2 ððs þ 3Þ2 þ 4Þ2
Therefore,
(
)
(
)
s
þ
3
s
þ
3
L1
¼ L1
ðs2 þ 6s þ 13Þ2
ððs þ 3Þ2 þ 4Þ2
(
)
s
3t 1
¼e L
:
ðs2 þ 4Þ2
n
o
s
To find L1 ðs2 þ4Þ
2 , we note that
d
1
2s
¼
:
2
ds s2 þ 4
ðs þ 4Þ2
Therefore,(
)
2s
1
1
L
¼ tL1 2
s þ4
ðs2 þ 4Þ2
and so
Proof: Since,
t
¼ sin 2t
2
(
)
s
1
L1
¼ t sin 2t:
2
2
4
ðs þ 4Þ
Consequently,
we get
(
)
sþ3
1
1
¼ t e3t sin 2t:
L
2
4
ðs2 þ 6s þ 13Þ
Solution. By Theorem 6.7, we have
n
o
8 1
9
= L1 1 1
<Z 1
u1
uþ1
1
du ¼
:
L1
:
;
u1 uþ1
t
s
But
L1
1
1
u1 uþ1
¼ et et :
Therefore,
8 1
9
=
<Z 1
1
et et
:
du ¼
L1
:
;
t
u1 uþ1
s
Theorem 6.8. (Multiplication by sn). If L1 fF ðsÞg ¼
f ðtÞ and f(0) ¼ 0, then
L1 fsF ðsÞg ¼ f 0 ðtÞ:
Proof: We know that
Lf f 0 ðtÞg ¼ s FðsÞ f ð0Þ
and so
L1 fsF ðsÞ f ð0Þg ¼ f 0 ðtÞ;
that is,
L1 fsF ðsÞg ¼ f 0 ðtÞ:
Remark 6.2. If f (0) 6¼ 0, then
Theorem 6.7. (Inverse Laplace Transform of
Integrals). If L1
ðsÞg ¼ f ð9
tÞ, then
8fF
<Z1
= f ðtÞ
FðuÞ du ¼
:
L1
:
;
t
s
L1 fsF ðsÞ f ð0Þg ¼ f 0 ðtÞ
and so
L1 fsF ðsÞg ¼ f 0 ðtÞ þ f ð0ÞðtÞ;
where (t) is the Dirac delta function.
Inverse Laplace Transform
n
6.9
Also g(0) ¼ g0 (0) ¼ 0. Thus
Lfg00 ðtÞg ¼ s2 LfgðtÞg s gð0Þ g0 ð0Þ
EXAMPLE 6.28
n 2 o
Find L1 s2sþ1 .
¼ s2 LfgðtÞg;
Solution. We know that
s
L1 2
s þ1
that is,
Lff ðtÞg ¼ s2 L fgðtÞg
¼ cos t ¼ f ðtÞ
and f(0) ¼ 1 6¼ 0 for t ¼ 0. Therefore,
s2
L1 2
¼ L1 fsF ðsÞg
s þ1
or
FðsÞ ¼ s2 L fgðtÞg
or
LfgðtÞg ¼
0
¼ f ðtÞ þ f ð0ÞðtÞ ¼ ðtÞ sin t:
Hence
EXAMPLE 6.29
n 3 o
Find L1 s2sþ1 .
L
1
FðsÞ
s2
Z t Zv
FðsÞ
s2
¼ gðtÞ ¼
f ðuÞ du dv
0
Solution. From the Example 6.28,
s2
¼ ðtÞ sin t ¼ f ðtÞ:
L1 2
s þ1
Since f (0) ¼ 0, we have
s3
L1 2
¼ L1 fsF ðsÞg ¼ f 0 ðtÞ
s þ1
d
¼ fðtÞ sin tÞ ¼ 0 ðtÞ cos t:
dt
Theorem 6.9. (Division by s). If L1 fF ðsÞg ¼ f ðtÞ,
then
Zt
1 FðsÞ
¼ f ðuÞ du:
L
s
0
Proof: We know that
8 t
9
<Z
= FðsÞ
f ðuÞ du ¼
L
:
:
;
s
0
Zt Zt
¼
f ðtÞ dt2 :
0
0
In general, we have
L
1
Zt Zt
Zt
FðsÞ
¼
. . . f ðtÞdtn :
sn
0 0
0
|fflfflfflfflfflfflffl{zfflfflfflfflfflfflffl}
n times
EXAMPLE 6.30
n
o
1
Find L1 sðsþ1Þ
.
Solution. Since L1
L1
n
1
sðs þ 1Þ
o
¼ et , we have
Zt
¼ eu du ¼ ½eu t0
1
sþ1
0
¼ ½et 1 ¼ 1 et :
0
Therefore,
L1
Zt
FðsÞ
s
f ðuÞdu:
¼
0
Remark 6.3. Consider
Z t Zv
gðtÞ ¼
f ðuÞ du dv:
0
Then
0
0
Zt
g ðtÞ ¼
f ðuÞ du
0
and
g00 ðtÞ ¼ f ðtÞ:
EXAMPLE 6.31
n
o
1
.
Find L1 s2 ðsþ1Þ
Solution. From Example 6.30, we have
1
¼ 1 et
L1
sðs þ 1Þ
Therefore, by Theorem 6.9, we have
Zt
1
1
¼ ð1 eu Þdu ¼ ½u þ eu t0
L
s2 ðs þ 1Þ
0
¼ t þ et 1:
6.10
n
Engineering Mathematics-II
EXAMPLE 6.32
n
o
n
o
s
1
Find L1 ðs2 þa
and deduce L1 ðs2 þa
2 Þ2
2 Þ2
from it.
Solution. Since
d
a
2as
¼
;
ds s2 þ a2
ðs2 þ a2 Þ2
we have
(
L1 2as
)
¼ tf ðtÞ;
þ a2 Þ 2
n
o
a
where f ðtÞ ¼ L1 s2 þa
¼ sin at and so
2
(
)
s
1
¼ t sin at
2a L
ðs2 þ a2 Þ2
(
Hence
L1
ðs2
s
ðs2 þ a2 Þ2
)
¼
t sin at
2a
Zt
where P(s) is a polynomial of degree less than n. In
terms of complex analysis, we call a1, a2, . . ., an the
simple poles of F(s). The partial fraction decomposition shall be
A1
A2
An
þ
þ ... þ
:
FðsÞ ¼
s a1 s a2
s an
Multiplying both sides of this equation by s ai and
letting s ! ai, we have
Ai ¼ lim ðs ai ÞFðsÞ:
Again, in terms of complex analysis, Ai is called the
residue of F(s) at the poles ai.
Now, let L{f(t)} ¼ F(s). Then
X
n
n
X
Ai
¼
L1
A i eai t
f ðtÞ ¼ L1 fFðsÞg ¼
s
a
i
i¼1
i¼1
¼
u sin au du
n
X
i¼1
0
9
8
Zt
=
<
h
i
t
1
cos au
cos au
u
du
¼
;
2a :
a
a
0
0
1
t cos at sin at
¼
þ 2
2a
a
a
1
¼ 3 ðsin at at cos atÞ:
2a
6.3
ai 6¼ aj ;
s!ai
Now Theorem 6.9 yields
(
)
1
1
L
ðs2 þ a2 Þ2
1
¼
2a
We can also use the concept of simple poles of
the rational function F(s) to find inverse Laplace
transform. Let
PðsÞ
PðsÞ
;
¼
FðsÞ ¼
QðsÞ ðs a1 Þðs a2 Þ . . . ðs an Þ
PARTIAL FRACTIONS METHOD TO FIND
INVERSE LAPLACE TRANSFORM
While working with physical problems, we come
across some functions that are not immediately
recognizable as the Laplace transform of some
elementary functions. In such cases, we decompose
the given function into partial fractions and then
write down the inverse of each fraction. This
method is used when inverse transform of a rational
function is required.
lim ðs ai ÞFðsÞeai t :
s!ai
EXAMPLE 6.33
n
o
Find L1 s3 ðs12 þ1Þ .
Solution. Write
1
A B C Ds þ E
;
¼ þ 2þ 3þ 2
s3 ðs2 þ 1Þ
s s
s
s þ1
which yields
1 ¼ As2 ðs2 þ 1Þ þ Bsðs2 þ 1Þ
þ Cðs2 þ 1Þ þ ðDs þ EÞs3 :
Putting s ¼ 0, we get C ¼ 1. Comparing coefficients
of s on both sides, we get B ¼ 0. Comparing
coefficients of s2 on both sides, we get A þ C ¼ 0
which yields A ¼ C ¼ 1. Comparing coefficients of s3, we get B þ E ¼ 0 and so E ¼ B ¼ 0.
Comparing coefficients of x4, we have A þ D ¼ 0
and so D ¼ A ¼ 1. Thus
1
1 1
s
¼ þ 3þ 2
:
s3 ðs2 þ 1Þ
s s
s þ1
Inverse Laplace Transform
Hence, by Example 6.1,
1
1 1
s
¼ L1 þ 3 þ 2
L1 3 2
s ðs þ 1Þ
s s
s þ1
1
¼ 1 þ t2 þ cos t
2
1 2
¼ ðt þ 2 cos t 2Þ:
2
6.11
EXAMPLE 6.36
n
o
Find L1 ðs1Þ4sþ5
.
2
ðsþ2Þ
Solution. Write
4s þ 5
¼
ðs 1Þ2 ðs þ 2Þ
A
B
C
þ
þ
;
s 1 ðs 1Þ2 s þ 2
which gives
4s þ 5 ¼ Aðs þ 2Þðs 1Þ þ Bðs þ 2Þ þ Cðs 1Þ2 :
EXAMPLE 6.34
n
o
Find L1 ðsþ1Þ 2ðs2 þ1Þ .
Solution. Write
2
A
Bs þ C
¼
:
þ 2
2
ðs þ 1Þ ðs þ 1Þ s þ 1 s þ 1
Clearing fractions, we have
Putting s ¼ 1, we get 9 ¼ 3B and so B ¼ 3. Equating
the coefficients of s and s2, we get A þ B – 2C ¼ 4
and A þ C ¼ 0. These equations give C ¼ 13 and
A ¼ 13. Thus
4x þ 5
2
2 ¼ Aðs2 þ 1Þ þ ðBs þ CÞðs þ 1Þ:
Putting s ¼ 1, we have A ¼ 1. Comparing the
coefficient of s2 and that of s0, we get
0 ¼ A þ B; which yields B ¼ 1
and
2 ¼ A þ C; which yields C ¼ 1:
Hence
2
1
s þ 1
¼
þ
ðs þ 1Þðs2 þ 1Þ s þ 1 s2 þ 1
1
s
1
þ
:
¼
s þ 1 s2 þ 1 s 2 þ 1
Therefore,
2
¼ et cos t þ sin t:
L1
ðs þ 1Þ ðs2 þ 1Þ
ðs 1Þ ðs þ 2Þ
Hence
(
L
1
1
3
1
þ
:
2
3ðs 1Þ ðs 1Þ
3ðs þ 2Þ
¼
)
4s þ 5
2
ðs 1Þ ðs þ 2Þ
1
1
¼ et þ 3t et e2t :
3
3
EXAMPLE 6.37
n
o
2
Find L1 s3 þ6s2sþ11sþ6 .
Solution. We have
s2
:
s3 þ 6s2 þ 11s þ 6
Clearly s ¼ 1, 2, and 3 are roots of the polynomial s3 þ 6s2 þ 11s þ 10. Hence the simple poles
of F(s) are 1, 2, and 3. Therefore,
FðsÞ ¼
f ðtÞ ¼ L1 fFðsÞg ¼ lim ðs þ 1Þ FðsÞet
s!1
þ lim ðs þ 2Þ FðsÞe2t þ lim ðs þ 3Þ FðsÞe3t
EXAMPLE 6.35
n
o
s2
Find L1 ðs1Þ ðsþ2Þ
ðs3Þ .
s!2
Solution. The simple poles of the rational function
F(s) are s ¼ 1, 2, and 3. Therefore,
1
f ðtÞ ¼ L fFðsÞg ¼ limðs 1Þ FðsÞe
t
s!1
þ lim ðs þ 2Þ FðsÞ e2t þ lim ðs 3Þ FðsÞ e3t
s!2
n
s!3
1
4
9
¼ et þ
e2t þ
e3t
6
ð3Þ ð5Þ
ð2Þ ð5Þ
1
4
9
¼ et þ e2t þ e3t :
6
15
10
s!3
1
9
¼ et 4e2t þ e3t :
2
2
EXAMPLE 6.38
n 3 o
s
Find L1 s4 a
.
4
Solution. We have
s3
s3
¼
s4 a4 ðs aÞ ðs þ aÞ ðs2 þ a2 Þ
¼
A
B
Cs þ D
þ
þ 2
;
s a s þ a s þ a2
6.12
n
Engineering Mathematics-II
which yields
s3 ¼ Aðs þ aÞðs2 þ a2 Þ þ Bðs aÞðs2 þ a2 Þ
¼
þ ðCs þ DÞðs2 a2 Þ
¼ Aðs þ as þ a s þ a Þ þ Bðs as þ a s a Þ
3
2
2
3
3
2
2
3
þ Cðs3 sa2 Þ þ Dðs2 a2 Þ:
Putting s ¼ a, a, we get A ¼ B ¼ 14. Comparing
coefficients of s, and s2, we get C ¼ 12, D ¼ 0.
Thus 3
s
1
1
s
:
¼
þ
þ 2
4
4
s a
4ðs aÞ 4ðs þ aÞ 2ðs þ a2 Þ
Hence
s3
1
1
1
L1 4
¼ eat þ eat þ cos at
s a4
4
4
2
1
1 eat þ eat
¼ cos at þ
2
2
2
1
¼ ½cos at þ cosh at:
2
EXAMPLE 6.39
n
o
s2
Find L1 ðs2 a2 Þ ðs2 b
2 Þ ðs2 c2 Þ .
a ðeat eat Þ
b ðebt ebt Þ
þ 2 2 2 2
2
2
2
2
2ða b Þ ða c Þ 2ðb a Þ ðb c Þ
þ
þ
cðec t ec t Þ
asinhat
¼
2ðc2 a2 Þ ðc2 b2 Þ ða2 b2 Þ ða2 c2 Þ
bsinhbt
ðb2 a2 Þ ðb2 c2 Þ
þ
csinhct
ðc2 a2 Þ ðc2 b2 Þ
:
EXAMPLE 6.40
n 2
o
2s 6sþ5
Find L1 s3 6s
2 þ11s6 .
Solution. Let
FðsÞ ¼
s3
2s2 6s þ 5
:
6s2 þ 11s 6
We observe that s ¼ 1, s ¼ 2, and s ¼ 3 are the roots
of the equation s3 6s2 þ 11s 6 ¼ 0. Thus
FðsÞ ¼
2s2 6s þ 5
:
ðs 1Þ ðs 2Þ ðs 3Þ
Solution. Let
Therefore,
s2
FðsÞ ¼ 2
2
2
2
2
2
ðs a Þ ðs b Þ ðs c Þ
L1 fFðsÞg ¼ limðs 1Þ FðsÞ et þ limðs 2Þ FðsÞe2t
s!1
s!2
2
s
:
¼
1
5
ðs aÞ ðs þ aÞ ðs bÞ ðs þ bÞðs cÞ ðs þ cÞ
þ limðs 3Þ FðsÞ e3t ¼ et e2t þ e3t :
s!3
2
2
The simple poles of F(s) are a, a, b, b, c, and c.
Therefore,
EXAMPLE 6.41
L1 fFðsÞg ¼ limðsaÞFðsÞeat þ lim ðsþaÞFðsÞ eat
1 s!a
s!a
Find L1 s3 a
3 .
bt
bt
þlimðsbÞFðsÞe þ lim ðsþbÞFðsÞe
s!b
s!b
þlimðscÞFðsÞect þ lim ðsþcÞFðsÞect
s!c
s!c
a2
eat
2aða2 b2 Þða2 c2 Þ
a2
eat
2aða2 b2 Þ ða2 c2 Þ
b2
ebt
þ
2bðb2 a2 Þðb2 c2 Þ
b2
ebt
2bðb2 a2 Þ ðb2 c2 Þ
c2
ec t
þ
2
2cðc a2 Þ ðc2 b2 Þ
c2
ec t
2
2cðc a2 Þðc2 b2 Þ
¼
Solution. We have
1
1
FðsÞ ¼ 3
¼
s a3 ðs aÞ ðs2 þ as þ a2 Þ
¼
A
Bs þ C
þ
:
s a s2 þ as þ a2
Therefore,
1 ¼ Aðs2 þ as þ a2 Þ þ ðBs þ CÞðs aÞ:
Putting s ¼ a, we have A ¼ 3a12 . Comparing
the coefficients of constant terms and of s2, we get
respectively, a2A aC ¼ 1 and A þ B ¼ 0 which in
2
. Thus
turn imply B ¼ A ¼ 3a12 and C ¼ 3a
1
1
s þ 2a
2 2
FðsÞ ¼ 2
:
3a ðs aÞ 3a s þ as þ a2
Inverse Laplace Transform
Hence
L1 fFðsÞg
a 3a !
sþ2 þ 2
1 1 1
1 1
2L
¼ 2L
a2 3a2
3a
sa
3a
sþ2 þ 4
"
!
a
sþ2
1 at
1
1
¼ 2e 2 L
a2 3a2
3a
3a
sþ þ 4
!# 2
1
þL
3a
2
a2 3a2
sþ2 þ 4
pffiffiffi pffiffiffi
1 at
1
3 a pffiffiffi
3a
at2
cos
¼ 2e 2 e
t þ 3 sin
t
3a
3a
2
2
pffiffiffi pffiffiffi
pffiffiffi
1
3a
3a
at
t þ 3 sin
t :
¼ 2 eat e 2 cos
3a
2
2
EXAMPLE 6.42
n
o
4sþ5
Find L1 ðs1Þ
.
2
ðsþ2Þ
n
6.13
Putting s ¼ 1 yields D ¼ 13. Putting s ¼ 2, we
have B ¼ 13. Comparing coefficients of s3, we have
A þ C ¼ 0 and comparing coefficient of s2, we have
B þ D 3C ¼ 0. Thus C ¼ A ¼ 0 and so
2s 1
1
1
¼
þ
:
FðsÞ ¼
2
2
2
ðs þ 2Þ ðs 1Þ
3ðs þ 2Þ
3ðs 1Þ2
Hence
1
1
1
L1 fFðsÞg ¼ t e2t þ t et ¼ tðet e2t Þ:
3
3
3
6.4
HEAVISIDE’S EXPANSION THEOREM
PðsÞ
Let F(s) ¼ QðsÞ
, where P(s) and Q(s) have no
common factors and the degree of Q(s) is greater
than the degree of P(s). If Q(s) has distinct zeros,
then the following theorem gives us the inverse
Laplace transform of F(s).
Theorem 6.10. (Heaviside’s Expansion Formula). Let
P(s) and Q(s) be polynomials in s, where degree of
Solution. Write
4s þ 5
A
B
C
Q(s) is greater than that of P(s). If Q(s) has n distinct
þ
;
þ
FðsÞ ¼
¼
ðs 1Þ2 ðs þ 2Þ s 1 ðs 1Þ2 s þ 2 zeros a1, a2, . . ., an, then
n
X
Pðai Þ ai t
and so
1 PðsÞ
e :
L
¼
0 ða Þ
4s þ 5 ¼ Aðs 1Þðs þ 2Þ þ Bðs þ 2Þ þ Cðs 1Þ2 :
QðsÞ
Q
i
i¼1
Taking s ¼ 1, we get B ¼ 3. Similarly, taking
s ¼ 2, we get C ¼ 13. Now equating coefficients
of s2, we have A þ C ¼ 0 and so A ¼ C ¼ 13.
Hence
1
3
1
þ
:
FðsÞ ¼
3ðs 1Þ ðs 1Þ2 3ðs þ 2Þ
Hence
1
1
L1 fFðsÞg ¼ et þ 3t et e2t :
3
3
EXAMPLE 6.43
n
o
Find L1 ðsþ2Þ2sþ1
.
2
ðs1Þ2
Proof: Since Q(s) has n distinct zeros, it factorizes
into n linear factors and so by partial fractions, we
have
PðsÞ
A1
A2
Ai
An
¼
þ
þ ... þ
þ ... þ
:
s ai
s an
QðsÞ s a1 s a2
Multiplying throughout by s ai and letting s ! ai,
we obtain
PðsÞ
s ai
Ai ¼ lim
ðs ai Þ ¼ lim PðsÞ
s!ai QðsÞ
s!ai
QðsÞ
s ai
¼ lim PðsÞ lim
s!ai
s!ai QðsÞ
Solution. Write
FðsÞ ¼
¼ Pðai Þ lim
2s þ 1
2
2
ðs þ 2Þ ðs 1Þ
A
B
C
D
¼
þ
þ
þ
2
s þ 2 ðs þ 2Þ
s 1 ðs 1Þ2
and so
2s þ 1 ¼ Aðs þ 2Þðs 1Þ2 þ Bðs 1Þ2
þ Cðs 1Þðs þ 2Þ2 þ Dðs þ 2Þ2 :
1
s!ai Q0 ðsÞ
¼
Hence
ðusing L’Hospital’s
Pðai Þ
:
Q0 ðai Þ
n
PðsÞ X
PðQi Þ
1
¼
0 ða Þ ðs a Þ
QðsÞ
Q
i
i
i¼1
ruleÞ
6.14
n
Engineering Mathematics-II
and so
PðsÞ
L1
QðsÞ
6.5
n
X
Pðai Þ 1
1
L
¼
0 ða Þ
Q
s
ai
i
i¼1
n
X Pðai Þ
eai t :
¼
0 ða Þ
Q
i
i¼1
EXAMPLE 6.44
n
o
Find L1 ðs3Þ sðs2 þ4Þ using Heaviside’s expansion
formula.
Solution. Let
FðsÞ ¼
PðsÞ
s
:
¼
QðsÞ ðs 3Þ ðs2 þ 4Þ
SERIES METHOD TO DETERMINE INVERSE
LAPLACE TRANSFORM
Let L{f(t)} ¼ F(s). In certain situation, we observed
that Laplace transform of f(t) can be obtained by
expressing f(t) in terms of power series and then by
taking transform term-by-term. The same technique
proves useful in finding inverse Laplace transforms.
Thus, if
a0 a1 a2
FðsÞ ¼ þ 2 þ 3 þ . . . ;
s
s
s
then
f ðtÞ ¼ a0 þ a1 þ a2
Thus
PðsÞ ¼ s; QðsÞ ¼ ðs 3Þðs2 þ 4Þ
¼ s3 3s2 þ 4s 12
and
where the series on the right-hand side may be summable to a known function.
Q0 ðsÞ ¼ 3s2 6s þ 4:
Also roots of Q(s) ¼ 0 are 3, 2i, and 2i. Therefore,
by Heaviside’s expansion formula
3
X
PðsÞ
Pðai Þ ai t Pð3Þ 3t
e ¼ 0 e
¼
L1
QðsÞ
Q0 ðai Þ
Q ð3Þ
i¼1
Pð2iÞ 2it Pð2iÞ 2it
e þ 0
e
Q0 ð2iÞ
Q ð2iÞ
3
2i
2i
¼ e3t þ
e2it e2it
13
8 12i
8 þ 12i
3
ð3 þ 2iÞ
ðcos 2t þ isin 2tÞ
¼ e3t 13
26
2i 3
þ
ðcos 2t i sin 2tÞ
26
3
3
2
¼ e3t cos 2t þ sin 2t:
13
13
13
þ
EXAMPLE 6.45
n
o
3sþ7
Find L1 ðs3Þ
ðsþ1Þ using Heaviside’s expansion
formula.
Solution. We have
PðsÞ
3s þ 7
3s þ 7
:
¼
¼
FðsÞ ¼
QðsÞ ðs 3Þ ðs þ 1Þ s2 2s 3
0
Here P(s) ¼ 3s þ 7, Q(s) ¼ s 2s 3, Q (s) ¼
2s 2, and zeros of Q(s) are 3 and 1. Hence, by
Heaviside’s expansion formula,
2
X
Pðai Þ ai t Pð3Þ 3t Pð1Þ t
e ¼ 0 e þ 0
e
L1 fFðsÞg ¼
Q0 ðai Þ
Q ð3Þ
Q ð1Þ
i¼1
2
¼
16 3t 4 t
e þ e ¼ 4e3t et :
4
4
t2
þ ...;
2!
EXAMPLE 6.46
Find L1
e1=s
.
s
Solution. Expansion in series yields
1 1=s 1
1
1
1
1
¼ 1 þ
þ
...
e
2
3
s
s
s 2! s
3! s
4! s4
1 1
1
1
þ ...
¼ 2þ
s s
2! s3 3!s4
n
1
X ð1Þ
:
¼
n! snþ1
n¼0
Inversion of the series term-by-term yields
1
X
ð1Þn 1 1
L
L1 fFðsÞg ¼
n!
snþ1
n¼0
¼
1
X
ð1Þn tn
n¼0
ðn!Þ
2
pffi
¼ J0 ð2 tÞ;
where J0(t) is the Bessel function of order zero.
EXAMPLE 6.47
Find L1 1s sin 1s .
Solution. We have
"
#
1
1 1 1 ð1=sÞ3 ð1=sÞ5 ð1=sÞ7
þ ...
sin ¼
3!
5!
7!
s
s s s
¼
1
1
1
1
4 þ 6 8 þ ...
2
s
s :3! s :5! s 7!
Inverse Laplace Transform
Therefore,
1 1
1
1
1
¼ L1 2 L1 4
L1 sin
s
s
s
3!
s
þ
1 1 1
1
1
L1 8 þ .. .
L
6
5!
s
7!
s
¼t
1
ð3!Þ2
t3 1
ð5!Þ2
t5 1
ð7!Þ2
t7 þ ...
n
6.15
But, using G(vþ1) ¼ vG(v), we have
1
1
1:3:5 . . . :ð2n 1Þ
nþ
¼
2
2
2n
pffiffiffi 1:3:5 . . . :ð2n 1Þ
¼ :
2n
Hence
1
1 X
ð1Þn an tn
1
pffiffiffi
L1 fFðsÞg ¼ pffi
¼ pffiffiffiffiffi eat :
n!
t n¼0
t
EXAMPLE 6.48
Find L1 tan1 1s .
6.6
Solution. We have
1=s ð1=sÞ3 ð1=sÞ5 ð1=sÞ7
1 1
tan
þ
þ ...
¼
3
5
7
s
1
The convolution of two functions plays an important
role in a number of physical applications. It is
generally expedient to resolve a Laplace transform
into the product of two transforms when the inverse
transform of both transforms are known.
1
1
1
1
¼ 3 þ 5 7 þ ...
s 3s
5s
7s
Therefore, inversion of the series term-by-term
yields
1 t2 1 t4 1 t6
1
1 1
tan
¼ 1 : þ : : þ ...
L
s
3 2! 5 4! 7 6!
CONVOLUTION THEOREM
Definition 6.2. The convolution (or Faltung) of two
given functions f(t) and g(t), t > 0 is defined by the
integral
Zt
ðf gÞðtÞ ¼ f ðuÞgðt uÞ du;
0
which, of course, exists if f and g are piecewise
t2 t4 t6
continuous.
¼ 1 þ þ ...
Rt
3! 5! 7!
The integral 0 f ðuÞ gðt uÞ du represents a
1
t3 t5 t7
sint superposition of effects of magnitude f (u) occur:
¼ t þ þ ... ¼
t
t
3! 5! 7!
ring at time t ¼ u for which g (t – u) is the influence
function or response of a system to a unit impulse
defined at time t ¼ u.
EXAMPLE 6.49
n
o
1 pffiffiffiffiffiffi
1
If we substitute v ¼ t – u, then
Find L
.
sþa
Zt
ðf gÞðtÞ ¼ gðvÞ f ðt vÞdv ¼ ðg f ÞðtÞ:
Solution. Let
0
1
1
a 1=2
FðsÞ ¼ pffiffiffiffiffiffiffiffiffiffi ¼ pffiffi 1 þ
Hence,
the
convolution
is commutative. Also, it
s
s
sþa
follows
from
the
definition
that
1
1 a 12 : 32 a2 12 : 32 : 52 a3
aðf gÞ ¼ af g ¼ f ag; a constant
¼ pffiffi 1 þ...
þ
2! s
3! s
2 s
s
f ðg þ hÞ ¼ ðf gÞ þ ðf hÞ ðdistributive propertyÞ:
n
1
X ð1Þ 1:3:5 . . . ð2n 1Þan
Further, we note that
jsj > jaj:
¼
n n! snþð1=2Þ
2
½f ðg hÞðtÞ
n¼0
Zt
Hence
n
1
n nð1=2Þ
X
¼ f ðuÞðg hÞðt uÞdu
ð1Þ 1:3:5 . . . ð2n 1Þa t
L1 fFðsÞg ¼
n n!ðn þ 1=2Þ
0
2
0 tu
1
n¼0
Zt
Z
n
1
n
n
1 X ð1Þ 1:3:5 . . . :ð2n 1Þa t
¼ f ðuÞ@ gðvÞhðt u vÞ dvA du
:
¼ pffi
n
2 n! ðn þ 1=2Þ
t n¼0
0
0
6.16
n
Engineering Mathematics-II
Zt
¼
Zt
f ðuÞ
0
Zt
¼
0
¼ ½ðf
Hence
0
@
Zz
gðz uÞ hðt zÞdzÞdu;
v¼zu
1
u
Proof: Using the definition of the Laplace transform,
Zt
Z1
we have
st
e
f ðuÞ gðt uÞ du dt
Lfðf gÞðtÞg ¼
0
Z1 Z t
f ðuÞgðz uÞ duAhðt zÞ dz
¼
0
gÞ hðtÞ:
0
f ðg hÞ ¼ ðf gÞ h;
and so convolution is associative.
EXAMPLE 6.50
Evaluate cos t
sin t.
0
est f ðuÞ gðt uÞdu dt:
0
The domain of this double integral is an infinite
wedge bounded by t ¼ 0, t ¼ 1, u ¼ 0, and u ¼ t
as displayed in Figure 6.1. Due to hypotheses on
f and g, the double integral on the right hand side
converges absolutely and so we can perform change
of order of integration.
u
Solution. By definition,
Zt
cos t sin t ¼ cosu sinðt uÞdu
u=t
0
Zt
¼
1
½sin t þ sinðt 2uÞ du
2
0
1
1
¼ sin t½ut0 þ ½cosðt 2uÞt0
2
4
1
1
1
¼ t sin t þ ½cosðtÞ cos t ¼ t sin t:
2
4
2
EXAMPLE 6.51
Evaluate et t.
Solution. By definition of convolution,
Zt
t
e t ¼ eu ðt uÞ du
t
0
Figure 6.1 Region of Integration
We take a strip parallel to the axis of t. Then t varies
from u to 1 and 1
u varies from 0 to 1. Hence
Z Z1
Lff ðtÞ ‘gðtÞg ¼
est f ðuÞgðt uÞ dt du
0
¼
0
Z1
0
¼ ½t eu t0 ½ðueu eu Þt0 ¼ et t 1:
Convolution is very significant in the sense that the
Laplace transform of the convolution of two functions is the product of their respective Laplace transforms. We prove this assertion in the form of the
following theorem.
u
Z1 Z1
¼
esu :esðtuÞ f ðuÞ gðt uÞ dt du
u
esu f ðuÞ
Z1
u
0
Implementing the change of variable t – u ¼ v in the
inner integral, we get1
Z1
Z
su
e f ðuÞ esv gðvÞdv du
Lf f ðtÞ gðtÞg ¼
0
Z1
0
Theorem 6.11. (Convolution Theorem). If f and g are
piecewise continuous on [0, 1) and of exponential
order c, then
¼
Lfð f gÞðtÞg ¼ Lf f ðtÞg:LfgðtÞg ¼ FðsÞGðsÞ;
or equivalently,
¼ GðsÞ
L1 fF ðsÞGðsÞg ¼ ð f
gÞðtÞ:
esðtuÞ gðt uÞdt du:
esu f ðuÞGðsÞdu
0
Z1
0
esu f ðuÞ du
¼ GðsÞFðsÞ ¼ FðsÞGðsÞ:
Inverse Laplace Transform
L1 fFðsÞGðsÞg ¼ f ðtÞ gðtÞ:
L1 fFðsÞGðsÞg ¼ f
EXAMPLE 6.52
Find L{eat ebt}.
¼
EXAMPLE 6.53
n
o
Find L1 ðsaÞ1ðsbÞ .
EXAMPLE 6.56
n
o
s
Use Convolution Theorem to find L1 ðs2 þa
.
2
2Þ
(see also Example 6.32).
Solution. Let
s
Solution. By Convolution Theorem
1
ðs aÞ ðs bÞ
¼ eat
ðs2 þ a2 Þ2
ebt
Then
e e
; a 6¼ b:
eut ebðtuÞ du ¼
ab
at
eu ðt uÞ du ¼ et t 1:
0
1
:
ebt g ¼ Lfeat g:Lfebt g ¼
ðs aÞ ðs bÞ
Zt
g ¼ t et
Zt
Solution. By Convolution Theorem,
L1
6.17
1
. Then, f (t) ¼ t
Solution. Let F(s) ¼ s12 and G(s) ¼ s1
t
and g(t) ¼ e . Therefore, by Convolution Theorem,
Consequently,
Lfeat
n
bt
¼
s
1
:
¼ FðsÞGðsÞ:
s2 þ a2 s2 þ a2
f ðtÞ ¼ L1 fF ðsÞg ¼ cos at
gðtÞ ¼ L1 fGðsÞg ¼
0
and
sin at
:
a
By Convolution Theorem,
Zt
sin aðt uÞ
1
L fF ðsÞGðsÞg ¼ f g ¼ cos au
du
a
EXAMPLE 6.54
n
o
Using Convolution Theorem, find L1 s3 ðs12 þ1Þ .
0
Solution. Let F(s) ¼ s13 , G(s) ¼ s2 1þ1 and so f(t) ¼ 12 t2
and g(t) ¼ sin t.
By Convolution Theorem and integration by
parts, we have
¼f
¼
8
1<
1
2a
½sin at þ sin aðt 2uÞ du
0
1
1
sin at½ut0 þ 2 ½cos aðt 2uÞt0
2a
4a
1
1
¼ t sin at þ 2 ½cos aðtÞ cos at
2a
4a
1
¼ t sin at:
2a
¼
L1 fFðsÞGðsÞg
1
g¼g f ¼
2
¼
Zt
Zt
sin uðt uÞ2 du
0
Zt
½ðt uÞ2 cos ut0 2ðt uÞ cos u du
9
=
; EXAMPLE 6.57
n
o
1
Using
Convolution
Theorem,
find
L
1
2
2
2
9
8
2
3
s ðs þa Þ .
Zt
=
<
1
¼ 4t2 2 ½ðt uÞ sin ut0 þ sin u du 5
Solution. Let
;
:
2
1
1
1
0
¼ 2: 2
¼ FðsÞGðsÞ:
2
2
2
s ðs þ a Þ s ðs þ a2 Þ
1 2
¼ ½t þ 2 cos t 2:
Then
2
f ðtÞ ¼ L1 fF ðsÞg ¼ t and
2:
0
n
o
EXAMPLE 6.55
1
Using Convolution Theorem, find L1 s2 ðs1Þ
.
gðtÞ ¼ L1 fGðsÞg ¼
sin at
:
a
6.18
n
Engineering Mathematics-II
By Convolution Theorem,
Then
sin at
:
a
Therefore, by Convolution Theorem,
f ðtÞ ¼ gðtÞ ¼
1
L fF ðsÞGðsÞg
Zt
1
1
ðt uÞ sin au du
¼ sin at t ¼
a
a
0
9
8
Zt
t
=
<
1
1
cos au
ðt uÞ cos au du
¼
;
a: a
a
0
L1 fFðsÞGðsÞg
¼f
1 t
sin au
a a
a2 0
1
1
¼ 2 t sin at :
a
a
¼
¼
1 t sin at
2
a a
a
¼
s2 ðs
Then
þ 1Þ
2
¼
f ðtÞ ¼ L1 fF ðsÞg ¼ t
So, by Convolution Theorem, we have
L1 fFðsÞGðsÞg ¼ g f
Zt
Zt
u
¼ ðue Þ ðt uÞ du ¼ ðtu u2 Þ eu du
¼ ½ðtu u
Þeu t0 Zt
1
½cos að2u tÞ cos at du
2
0
and
g ðtÞ ¼ L1 fGðsÞg ¼ t et
2
1
a2
Zt
2 t
3
Zt
Z
1 4
¼ 2
cos að2u tÞdu cos atdu5
2a
1
1
:
¼ FðsÞGðsÞ:
s2 ðs þ 1Þ2
0
sin au sin aðt uÞdu
0
EXAMPLE 6.58
n
o
1
Using Convolution Theorem, find L1 s2 ðsþ1Þ
2 .
Solution. Let
1
Zt
0
0
t
1
a2
g¼
sin að2u tÞ
cos at½ut0
2a
0
¼
1
2a2
¼
1 sin at sin aðtÞ
t cos at
2a2 2a
2a
¼
1
½sin at at cos at:
2a3
EXAMPLE 6.60
n 2 o
a
Use Convolution Theorem to find L1 sðsþaÞ
.
2
Solution. Let
a2
1
a2
¼ :
¼ FðsÞGðsÞ:
s ðs þ aÞ2
sðs þ aÞ
2
0
ðt 2uÞeu du
0
t
Then
1
¼ 1:
f ðtÞ ¼ L1
s
Zt
Zt
n
o
a2
¼ ðt 2uÞ eu du ¼ ½ðt 2uÞeu t0 2 eu du To find g ðtÞ ¼ L1
, we note that
sðsþaÞ2
0
0
d
1
1
eu t
:
¼
¼ tet þ t 2
¼ tet þ t þ 2et 2
ds s þ a
ðs þ aÞ2
1
0
0
¼ ðt þ 2Þet þ t 2:
EXAMPLE 6.59
n
o
1
Using Convolution Theorem, find L1 ðs2 þa
.
2 Þ2
Solution. Let
1
ðs2 þ
a2 Þ 2
Therefore,
(
L
1
2
ðs þ aÞ
and so
¼
ðs2
1
1
:
¼ FðsÞGðsÞ:
þ a2 Þ ðs2 þ a2 Þ
)
1
(
L
1
¼ t L1
a2
ðs þ aÞ2
1
sþa
¼ t eat
)
¼ a2 t eat :
Inverse Laplace Transform
Application of Convolution Theorem yields
Zt
Zt
1
2 au
2
L fFðsÞGðsÞg ¼ a ue du ¼ a
u eau du
0
0
¼ a2
at
0
Let
f ðtÞ ¼ ta1 ; gðtÞ ¼ tb1 ; a; b > 0:
Then
Zt
ðf
au t
te
e
a
a2
6.19
Solution. We know that Beta function is defined by
Z1
bða; bÞ ¼ xa1 ð1 xÞb1 dx:
0
9
8
=
< u eau t Z t eau
du
¼ a2
;
: a 0
a
n
ua1 ðt uÞb1 du:
gÞðtÞ ¼
0
0
Putting u ¼ vt, we get
¼ at eat eat þ 1
ðf
¼ 1 eat ðat þ 1Þ:
gÞðtÞ ¼ t
Z1
va1 ð1 vÞb1 dv
aþb1
0
EXAMPLE 6.61
n
o
1
Find L1 pffisðs1Þ
using Convolution Theorem.
Solution. Let
1
1
1
pffiffi
¼ pffiffi :
¼ FðsÞGðsÞ:
sðs 1Þ
s s1
Then
1
f ðtÞ ¼ L1 fFðsÞg ¼ pffiffiffiffiffi
t
1
t
gðtÞ ¼ L fGðsÞg ¼ e :
Lftaþb1 bða; bÞg ¼ Lfta1 g Lftb1 g
¼
and
¼
0
t
1
e
pffiffiffiffiffiffi eðtuÞ du ¼ pffiffiffi
u
2et
¼ pffiffiffi
pffi
Zt
0
taþb1 bða; bÞ ¼ L1
2
¼ et pffiffiffi
Zt
0
u
e
pffiffiffi du
u
which implies
2
ev dv ¼ et erf
2
ðaÞ ðbÞ
saþb
¼ ðaÞ ðbÞL1
ev dv;u ¼ v2
pffi
Zt
ðaÞ ðbÞ ðaÞ ðbÞ
: b ¼
sa
s
saþb
and so
Hence Convolution Theorem yields
1
¼ L1 fFðsÞGðsÞg ¼ f g
L1 pffiffi
sðs þ 1Þ
Zt
¼ taþb1 bða; bÞ:
But by Convolution Theorem,
Lff gg ¼ Lff ðtÞgLfgðtÞg:
Therefore,
pffi
t;
0
pffi
where erf t is the error function.
EXAMPLE 6.62
Using Convolution Theorem, establish Euler’s formula for Beta function:
ðaÞ ðbÞ
bða; bÞ ¼
:
ða þ bÞ
¼ ðaÞ ðbÞ
1
saþb
taþb1
;
ða þ bÞ
ðaÞ ðbÞ
;
ða þ bÞ
the required Euler’s Formula for Beta function.
bða; bÞ ¼
Remark 8
6.4. It also follows from
9 Example 6.62 that
t
Z
= ðaÞ ðbÞ
<
ua1 ðt uÞb1 du ¼
:
L
;
:
saþb
0
EXAMPLE 6.63
Using Convolution Theorem, show that
Z1
2
cos tx
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx;
J0 ðtÞ ¼
1 x2
0
where J0 represents Bessel’s function of order zero.
6.20
n
Engineering Mathematics-II
6.7
Solution. We know that
1
1
1
LfJ0 ðtÞg ¼ pffiffiffiffiffiffiffiffiffiffiffiffi ¼ pffiffiffiffiffiffiffiffiffiffi : pffiffiffiffiffiffiffiffiffiffi
2
s
þ
i
si
s þ1
and so
1
1
J0 ðtÞ ¼ L1 pffiffiffiffiffiffiffiffiffiffi : pffiffiffiffiffiffiffiffiffiffi
sþi si
¼ L1 fFðsÞGðsÞg;
where
1
1
FðsÞ ¼ pffiffiffiffiffiffiffiffiffiffi ; GðsÞ ¼ pffiffiffiffiffiffiffiffiffiffi :
sþi
si
COMPLEX INVERSION FORMULA
The complex inversion formula is a technique for
computing directly the inverse of a Laplace transform. For this technique, we require the parameter
s to be a complex variable. The contour integration is the main tool applied in this technique.
Let f be a continuous function possessing a
Laplace transform and defined in (1, 1) such
that f (t) ¼ 0 for t < 0. Let s ¼ x þ iy be a complex
variable. Then
Z1
est f ðtÞ dt
Lff ðtÞg ¼ FðsÞ ¼
Since
L1
1
pffiffiffiffiffiffiffiffiffiffiffi
sþa
Z1
1
¼ pffiffiffiffiffi eat ;
t
¼
1
Z1
we have
1
f ðtÞ ¼ pffiffiffiffiffi eit
t
and
¼
1
gðtÞ ¼ pffiffiffiffiffi eit :
t
J0 ðtÞ ¼ L fFðsÞGðsÞg ¼ f
Zt
¼
0
1
¼
g
1
1
pffiffiffiffiffiffi eiu pffiffiffi
eiðtuÞ
1=2
u
ðt uÞ
Zt
!
du
eiðt2uÞ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi du:
uðt uÞ
0
Putting u ¼ tv, we get
Z1
1 eit ð1 2vÞ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi dv:
J0 ðtÞ ¼
vð1 vÞ
0
Now putting x ¼ 1 2v, we have
Z1
1
eitx
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx
J0 ðtÞ ¼
1 x2
1
1
¼
¼
2
Z1
1
Z1
0
cos tx
i
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx þ
2
1x
Z1
1
eðxþiyÞt f ðtÞ dt
eiyt ðext f ðtÞÞ dt:
1
Hence, by Convolution Theorem
1
0
sin tx
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx
1 x2
cos tx
pffiffiffiffiffiffiffiffiffiffiffiffiffi dx;
1 x2
the second integral vanishes because sin x is odd.
The integral on the right is the Fourier transform of
the function ext f ðtÞ. Thus the Laplace transform of
f (t) is the Fourier transform of the function ext f ðtÞ.
Suppose that f and its derivative f0 are piecewise
continuous on (1, 1), that is, both f and f0 are
continuous in any finite interval except possibly for
a finite number of jump discontinuities. Suppose,
further, that f is absolutely integrable, that is,
R1
jf ðtÞjdt < 1. Then Fourier integral theorem
1
asserts that
f ðtþÞ þ f ðtÞ
1
¼
2
2
Z1 Z1
f ðuÞ eisðtuÞ du ds;
1 1
where the integration with respect to s is in the
Cauchy principal value sense.
If f were continuous on (1, 1), then
f ðtþÞþf ðtÞ
converts to f(t) in the above expression.
2
Further, in terms of Fourier transform of f(t), the
above expression becomes
Z1
f ðtþÞ þ f ðtÞ
1
¼
FðsÞeist ds;
2
2
where F(s) ¼
of f(t).
R1
1
1
isu
e
f ðuÞdu is Fourier transform
Inverse Laplace Transform
Theorem 6.12. (Complex Inversion Formula). Let f be
continuous on [0, 1), f (t) ¼ 0 for t < 0, f be of
exponential order c and f0 be piecewise continuous
on [0, 1). If L{f (t)} ¼ F(s) and the real number c is
so chosen that all the singularities of F(s) lie in the
left of the line Re (s) > c, then
cþi1
Z
1
1
est FðsÞ ds; t > 0:
f ðtÞ ¼ L fFðsÞg ¼
2i
ci1
Proof: By definition of Laplace transform,
Z1
esu f ðuÞ du:
FðsÞ ¼
n
6.21
Remark 6.5. The expression
cþi1
Z
1
est FðsÞ ds
f ðtÞ ¼
2i
ci1
is called complex inversion formula, Bromwich
integral formula, Fourier–Mellin inversion formula,
or fundamental theorem of Laplace transform. In
practice, the integral in the above expression is
evaluated by considering the contour integration
along the contour GR, ABCDEA, known as
Bromwich contour shown in the Figure 6.2.
0
We have then
1
lim
T!1 2i
B
cþiT
cþiT Z1
Z
Z
1
st
e FðsÞds ¼ lim
est esu f ðuÞ du ds:
T!1 2i
ciT
ciT
R
0
Let s ¼ c þ iy so that ds ¼ idy. Therefore,
1
lim
T!1 2i
C
cþiT
Z
ZT
Z1
1
est FðsÞds ¼ lim ect
ei yt dy ei yu ecu f ðuÞdu
T!1 2
T
ciT
¼
1 ct
e
2
Z1 Z1
1 0
1
Figure 6.2 Bromwich Contour
The vertical line at c is known as Bromwich line.
Thus the contour GR consists of arc CR (ABCDE) of
radius R and centre at the origin and the Bromwich
line EA. Thus,
Z
Z
1
1
est FðsÞ ds ¼
est FðsÞ ds
2i
2i
R
CR
1 1
c t
þ
f ðtÞ for t > 0:
1
2i
Z
est FðsÞ ds:
EA
Hence
lim
E
γ−i T
D
ei yðtuÞ ðec u f ðuÞÞ du dy:
Thus ext f(t) is absolutely integrable and so Fourier
integral theorem asserts that
Z1 Z1
1
ðecu f ðuÞÞei yðtuÞ du dy
2
¼e
γ
0
0
The given hypothesis and Theorem 6.1 imply that
L{f (t)} converges absolutely for Re (s) > c, that is,
Z1
Z1
st
je f ðtÞjdt ¼
ext jf ðtÞjdt < 1; x > c:
0
γ+i T
A
1
T !1 2i
cþiT
Z
est FðsÞds ¼
ciT
1
:2ec t f ðtÞ ec t
2
¼ f ðtÞ; t > 0;
and so
1
f ðtÞ ¼
2i
R
cþi1
Z
est FðsÞ ds;
ci1
Since F(s) is analytic for Re(s) ¼ x > c, all singularities of F(s) must lie to the left of the Bromwich
line. Thus, by Cauchy residue theorem, we have
Z
n
X
1
est FðsÞ ds ¼
Res ðzk Þ;
2i
k¼1
t > 0:
where Res(zk) is the residue of the function at the
pole s ¼ zk. Since est 6¼ 0, multiplying F(s) by est
does not affect the status of the poles zk of F(s).
6.22
n
Engineering Mathematics-II
If we can show thatZ
est FðsÞ ds ¼ 0;
lim
R!1
CR
then letting R ! 1, we get
cþi
Z T
n
X
1
f ðtÞ ¼ lim
ets FðsÞ ds ¼
Resðzk Þ;
T !1 2i
k¼1
ci T
and so inverse function f can be determined.
The following theorem shows that Laplace
transform is one-to-one.
Theorem 6.13. Let f (t) and g(t) be two piecewise
smooth functions of exponential order and let F(s)
and G(s) be the Laplace transforms of f (t) and g(t)
respectively. If F(s) ¼ G(s) in a half-place Re(s) > c,
then f (t) ¼ g(t) at all points where f and g are
continuous.
Proof: Suppose f and g are continuous at t 2 R. By
complex inversion formula, we have
cþi
Z 1
1
est FðsÞ ds;
f ðtÞ ¼
2i
ci 1
1
gðtÞ ¼
2i
Solution. We have
FðsÞ ¼
and so
jFðsÞj s
;
a2
s2
jsj
jsj
:
js2 a2 j jsj2 jaj2
2
If |s| 2|a|, then |a|2 jsj4 and so |s|2 |a|2
and we have
4=3
:
jFðsÞj jsj
2
3
4 jsj
EXAMPLE 6.65
Find the Laplace transform of f (t) ¼ cosh at and
verify the inversion formula.
s
Solution. By Example 6.6, we have F(s) ¼ s2 a
2 . The
function F(s) is analytic except at poles s ¼ a and
s ¼ a. The Bromwich contour is shown in the
Figure 6.3.
B
ci 1
est FðsÞds ¼ 0; t > 0:
CR
satisfies growth restriction
est GðsÞ ds:
Theorem 6.14. Let for s on CR, F(s) satisfies the
growth restriction
M
jFðsÞj p for p > 0; all R > R0 :
jsj
Then
Z
R!1
s
s2 a2
cþi
Z 1
Since F(s) ¼ G(s), it follows that f (t) ¼ g(t). Thus
L{f (t)} ¼ L{g(t)} implies that f (t) ¼ g(t) and so
Laplace operator is one-to-one.
Generally, we see that most of the Laplace
transforms satisfy the growth restriction
M
jFðsÞj p
jsj
for all sufficiently large values of |s| and some p > 0.
Obviously, F(s) ! 0 as |s| ! 1. Therefore, the
following result (stated without proof) is helpful.
lim
EXAMPLE 6.64
Show that F(s) ¼
condition.
γ+iT
A
R
C
−a
0 γ
a
E
γ −i T
D
Figure 6.3
By inversion formula, we have
cþi1
Z
s s
1
1
st
L
¼
e
ds
s 2 þ a2
2i
s 2 a2
ci1
1
¼
2i
Z
R
sest
ds:
s 2 a2
Inverse Laplace Transform
Further, F(s) satisfies growth restriction condition.
Therefore, integral over contour CR (arc ABCDE)
tends to zero as R ! 1. Now
sest aeat
¼
;
s!a
s!a sþa
2a
sest eat
¼
ðaÞ:
ResðaÞ ¼ lim ðsþaÞest FðsÞ ¼ lim
s!ðaÞ
s!ðaÞ sa 2a
ResðaÞ ¼ limðsaÞ est FðsÞ ¼ lim
Hence
eat þ eat
L1 fFðsÞg ¼
¼ cosh at;
2
and so inversion formula is verified.
EXAMPLE 6.66
n
o
v
Find L1 s2 þv
2 , s > 0 using inversion formula.
Solution. We have
FðsÞ ¼
v
;s > 0
s2 þ v 2
The function F(s) has two simple poles at s ¼ ± iv.
By inversion formula
cþi
Z 1 1
v
1
ds
L fFðsÞg ¼
est 2
2i
s þ v2
ci 1
¼
v
2i
Z
R
s2
est
ds;
þ v2
where GR is the Bromwich contour shown in the
Figure 6.4:
B
iω
C
0
γ +i T
A
! 0 as R ! 1:
Hence
v
L1 fFðsÞg ¼
2iðsum of residue at
2i
eivt eivt
¼ sin vt:
¼
2i
ivÞ
EXAMPLE 6.67
n
o
Find L1 sðs2 1þa2 Þ using inversion formula.
Solution. We have
1
1
:
sðs2 þ
sðs iaÞ2 ðs þ iaÞ2
Thus F(s) has a simple pole at s ¼ 0 and a pole of
order 2 at s ¼ ± ia. Further, F(s) satisfies growth
restriction condition. Therefore, integral over the
contour CR goes to zero as R ! 1. Further,
est
1
Resð0Þ ¼ lim sest FðsÞ ¼ lim
¼ 4
s!0
s!0 ðs2 þ a2 Þ2
a
d
ResðiaÞ ¼ lim ðs iaÞ2 ets FðsÞ
s!ai ds
!
d
ets
it iat eiat
¼ lim
e 4;
¼
s!ai ds sðs þ iaÞ2
2a
4a3
FðsÞ ¼
a2 Þ 2
¼
d
ðs þ iaÞ2 ets FðsÞ
!
d
ets
it iat eiat
¼ lim
e
4:
¼
s!ia ds sðs iaÞ2
2a
4a3
ResðiaÞ ¼ lim
s!ia ds
γ
Hence
−i ω
1
it iat
1
þ
ðe eiat Þ 4 ðeiat þ eiat Þ
a4 4a3
2a
1
a
¼ 4 1 t sin at cos at :
a
2
f ðtÞ ¼
E
γ−i T
Figure 6.4
We have
eivt
;
2iv
e
eivt
ResðivÞ ¼ lim ðsþivÞest FðsÞ ¼ lim
¼
:
s!i v
s!iv siv 2iv
ResðivÞ ¼ lim ðsivÞest FðsÞ ¼ lim
s!iv
6.23
Further, let s ¼ c þ R eih, 2 h 3
2 . Then the
integral over the contour CR yields
Z
3=2 ect et Rðcos hþi sin hÞ R ei h
ec t R
ðc þ Rei h Þ2 þ v2 R2 c2 v2
=2
R
D
n
est
¼
s!iv sþiv
st
EXAMPLE 6.68
n
o
Find L1 ðsþ1Þ3 sðs1Þ2 using inversion formula.
Solution. We have
FðsÞ ¼
s
ðs þ 1Þ3 ðs 1Þ2
:
6.24
n
Engineering Mathematics-II
The function F(s) has poles of multiplicity 2 at s ¼ 1
and poles of multiplicity 3 at s ¼ 1. Residue at
s ¼ 1 is given by
"
#
d
ðs 1Þ2 sest
Resð1Þ ¼ lim
s!1 ds ðs þ 1Þ3 ðs 1Þ2
!
d
sest
¼ lim
s!1 ds ðs þ 1Þ3
1 t
¼
e ð2t 1Þ;
16
"
#
1 d2
ðs þ 1Þ3 sest
Resð1Þ ¼ lim
s!1 2! ds2 ðs þ 1Þ3 ðs 1Þ2
!
1 d2
sest
¼ lim
s!1 2 ds2 ðs 1Þ2
¼
1 t
e ð1 2t2 Þ:
16
The value of the integral over the contour CR tends
to zero as R ! 1. Hence
f ðtÞ ¼ sum of residues at the poles
1
1
¼ et ð1 2t2 Þ þ et ð2t 1Þ:
16
16
EXAMPLE 6.69
Derive Heaviside’s expansion formula using complex inversion formula.
PðsÞ
Solution. Let F(s) ¼ QðsÞ
, where P(s) and Q(s) are
polynomials having no common factors (roots) and
degree of Q(s) is greater than the degree of P(s).
Suppose Q(s) has simple zeros at z1, z2, . . ., zm. If
degree of P(s) and Q(s) are n and m, respectively,
then for a0 6¼ 0, bm 6¼ 0,
PðsÞ
an sn þ an1 sn1 þ . . . þ a0
¼
QðsÞ bm sm þ bm1 sm1 þ . . . þ b0
¼
a0
an þ anþ1
s þ . . . þ sn
:
b
b0
smn bm þ m1
s þ . . . þ sm
For sufficiently large |s|, we have
an1
a0
þ...þ n j jan jþjan1 jþ...þja0 j ¼ C1 say;
s
s
am1
b0
jbm1 j
jb0 j jbm j
jbm þ
... m
¼ C2 say;
þ...þ n j jbm j
jsj
jsj
2
s
s
jan þ
and so
PðsÞ C1 =C2
jFðsÞj ¼ :
QðsÞ jsjmn
Thus, F(s) satisfies growth restriction condition.
Further,
Resðzn Þ ¼ lim ðs zn Þest FðsÞ
s!zn
est PðsÞ
¼ lim QðsÞQðz Þ ;
s!zn
¼
n
since QðZn Þ ¼ 0
szn
ezn t Pðzn Þ
:
Q0 ðzn Þ
Hence, by inversion formula, we have
m
X
Pðzn Þ tzn
f ðzÞ ¼
e ;
0 ðz Þ
Q
n
n¼1
which is the required Heaviside’s expansion
formula.
EXAMPLE 6.70
1
Find L1 sð1þe
as Þ ,
formula.
using
complex inversion
1
Solution. Let F(s) ¼ sð1þe
as Þ. Then F(s) has a simple
pole at s ¼ 0. Further, 1 þ eas ¼ 0 yields
eas ¼ 1 ¼ eð2n1Þi ; n ¼ 0; 1; 2; . . .
and so sn ¼ 2n1
i, n ¼ 0, ±1, ±2, . . . are also
a
poles of F(s). Also, dsd ð1 þ eas Þs¼sn ¼ a 6¼ 0.
Therefore, sn are simple poles. Now
1
Resð0Þ ¼ lim s est FðsÞ ¼ ;
s!0
2
2n 1
Res
i ¼ lim ðs sn Þest FðsÞ
s!sn
a
ðs sn Þest 0
¼ lim
form
s!sn sð1 þ eas Þ 0
est þ t est ðs sn Þ
;by L’Hospital rule
a s eas þ 1 þ eas
i
et 2n1
etsn
a
¼
:
¼
ð2n 1Þ i
a sn easn
¼ lim
s!sn
Also, it can be shown that F(s) satisfies growth
restriction condition. Hence, by inversion formula,
Inverse Laplace Transform
at the points of continuity of f, we have
f ðtÞ ¼ sum of residues at the poles
1
X
2n1
1
1
etð a Þ i
¼ 2 n¼1 ð2n 1Þ i
1
1 2X
1
2n 1
¼ sin
t:
2 n¼1 ð2n 1Þ
a
Solution. (i) We first find partial fraction of
We have
EXAMPLE 6.71
pffi
Find L1 fea s g , a > 0.
Putting s ¼ 0, we get
Solution. We know
12) that
( (Exercise
pffi )
a s
a
1 e
p
ffi
;
L
¼ erfc
s
2 t
Taking s ¼ 4, we get
that is,
a
L erfc pffi
2 t
pffi
a s
¼ e s :
Therefore,
d
a
erfc pffi
L
dt
2 t
because erfc
Thus
L
Hence
affi
p
2 t
¼ s FðsÞ f ð0Þ
pffi
ea s
¼s
s
pffi
¼ ea s :
pffi
a
2
L1 fea s g ¼ pffiffiffiffiffiffiffi ea =ð4tÞ :
2 t3
3sþ1
s2 ðs2 þ4Þ.
3s þ 1
A B cs þ D
¼ þ þ
s2 ðs2 þ 4Þ s s2 s2 þ 4
or
3s þ 1 ¼ Asðs2 þ 4Þ þ Bðs2 þ 4Þ þ sðcs þ DÞ:
1
1 ¼ 4B which yields B ¼ :
4
11 ¼ 16C 4D
ð1Þ
Comparing coefficients of s, we get
3 ¼ 4A þ D
ð2Þ
Comparing coefficients of s2 ; we get
1
0 ¼ B þ C which yields C ¼ B ¼ :
4
Then (1) yields D ¼ 16cþ11
¼ 74. Now (2) gives
4
! 0 as t ! 0.
a
2
pffiffiffiffiffiffiffi ea =ð4tÞ
3
2 t
6.25
n
A¼
3 D 3 74
5
¼ :
¼
4
4
16
Hence
3ðs þ 1Þ
5
1
s
7
¼
þ
:
þ
s2 ðs2 þ 4Þ 16s 4s2 4ðs2 þ 4Þ 4ðs2 þ 4Þ
Therefore
6.8
MISCELLANEOUS EXAMPLES
EXAMPLE 6.72. Find L1
h
1
ðs3Þ2
i
:
1
Solution. Since Lfeat tg ¼ ðsaÞ
2 by shifting property,
we have
"
#
1
1
¼ te3t
L
2
ðs 3Þ
EXAMPLE 6.73. Find the inverse Laplace transforms of:
3s
and (ii) tan1 s22 .
(i) s23sþ1
ðs2 þ4Þ e
L1
3s þ 1
þ 4Þ
s2 ðs2
¼
5 1
1
7
þ t cos 2t þ : sin 2t:
16 4
4
8
Using second shifting property, we have
L1
for t > 3.
3s þ 1 3s
e
s2 ðs2 þ 4Þ
¼
5 1
þ ðt 3Þ
16 4
1
cos 2ðt 3Þ
4
7
þ sin 2ðt 3Þ
8
6.26
n
Engineering Mathematics-II
(ii). We note that
d
2
1
4
4s
¼ 4
tan1 2 ¼
4
3
ds
s
s þ4
1 þ s4 s
4s
¼
ðs2 þ 2Þ2 ð2sÞ2
4s
¼ 2
ðs þ 2 þ 2sÞðs2 þ 2 2sÞ
ðs2 þ 2s þ 2Þ ðs2 2s þ 2Þ
¼
ðs2 þ 2s þ 2Þðs2 2s þ 2Þ
1
1
¼ 2
þ 2
s 2s þ 2 s þ 2s þ 2
1
1
¼
þ
:
ðs 1Þ2 þ 1 ðs þ 1Þ2 þ 1
Therefore
(
L
1
1
2
ðs 1Þ þ 1
þ
)
1
2
ðs þ 1Þ þ 1
¼ t f ðtÞ;
that is,
Solution. Using linearity and shifting properties, we
have
se2 þ es
s2 þ 2
s
se2
es
1
þ
L
¼ L1 2
s þ 2
s2 þ 2
1
1
þ sin ðt 1ÞH ðt 1Þ
¼ cos t H t 2
2
1
¼ sin t H t sin t½H ðt 1Þ
2
1
H ð t 1Þ ;
¼ sin t H t 2
s
L1
where HðtÞ denotes Heavyside’s unit step function.
EXAMPLE 6.76. Find the inverse Laplace transform of
1
:
s2 ðs2 þ a2 Þ
et sin t þ et sin t ¼ t f ðtÞ:
Solution. We know that
Hence
f ðtÞ ¼
sin tðet et Þ 2 sin t sinh t
¼
:
t
t
Lfsin atg ¼
Therefore
EXAMPLE 6.74. Find the inverse Laplace transform of
ð5s þ 3Þ=ðs 1Þðs2 þ 2s þ 5Þ:
Solution. Using partial fractions, we have
5s þ 3
1
s þ 2
¼
þ 2
2
ðs 1Þðs þ 2s þ 5Þ s 1 s þ 2s þ 5
1
ðs þ 1Þ 3
¼
s 1 ðs þ 1Þ2 þ 4
1
sþ1
¼
s 1 ðs þ 1Þ2 þ 4
3
þ
ðs þ 1Þ2 þ 4
Therefore
L1
5s þ 3
ðs 1Þðs2 þ 2s þ 5Þ
¼ et et cos 2t þ 3et sin 2t:
EXAMPLE 6.75. Find inverse Laplace transform of
ðses=2 þ es Þ=ðs2 þ 2 Þ:
a
:
s 2 þ a2
L1
1
s 2 þ a2
Then
a
2
sðs þ a2
L1
¼
1
a
Z
t
1
¼ sin at:
a
sin at dt ¼
0
1 cos att
a
a
0
1
¼ 2 ½cos at 1
a
and
1
L
a
s2 ðs2 þ a2 Þ
1
¼ 2
a
Z
t
ðcos at 1Þdt
0
t
1 sin at
t
a2
a
0
1
t
¼ 3 sin at þ 2
a
a
1
1
¼ 2 t sin at :
a
a
¼
Note: This question can also be solved using
Convolution Theorem (see Example 6.57)
Inverse Laplace Transform
h
i
2
EXAMPLE 6.77. Find L1 cot1 sþ1
:
Solution. Since
d
2
1
cot
¼
ds
sþ1
¼
or
s2
f ðtÞ ¼ L1 fFðsÞg ¼ cos 2t and
1þ
1
2
2
sþ1
2
!
ðs þ 1Þ2
;
2
s2 þ 2s þ 5
2
þ 2s þ 5
(
L1
2
ðs þ 1Þ2 þ 4
¼ t f ðtÞ
gðtÞ ¼ L1 fGðsÞg ¼ cos 2t:
Therefore, by Convolution Theorem
Z t
1
cos 2u cos 2ðt uÞ du
L fFðsÞGðsÞg ¼ f g ¼
0
Z t
1
½cos 2t þ cos ð4u 2tÞ du
¼
2 0
1
sin ð4u 2tÞ
u cos 2t þ
2
4
1
sin 2t
¼ t cos 2t þ
2
4
1
¼ ½2t cos 2t þ sin 2t:
4
)
¼ t f ðtÞ
et sin 2t ¼ t f ðtÞ:
1
f ðtÞ ¼ et sin 2t:
t
EXAMPLE 6.78. Find:
L
1
s2 þ 1
log
:
s ð s þ 1Þ
Solution. Similar to Example 6.22. In fact,
d
s2 þ 1
log
ds
sðs þ 1Þ
d
¼ ½logðs2 þ 1Þ log s logðs þ 1Þ
ds
2s
1
1
¼ 2
:
s þ1 s sþ1
Therefore
2s
1
1
¼ tf ðtÞ
L1 2
s þ1 s sþ1
or
2 cos t 1 et ¼ tf ðtÞ:
Hence
1
f ðtÞ ¼ ½et þ 1 2 cos t:
t
EXAMPLE 6.79. Find the inverse Laplace transform
s2
of ðs2 þ4Þ
2 :
t
¼
or
Hence
6.27
s
Solution. Let FðsÞ ¼ s2 þ4
¼ GðsÞ. Then
we have
L1
n
0
EXAMPLE 6.80. Using Convolution Theorem find the
Laplace inverse of
2
ðs þ 2Þ= s2 þ 4s þ 13 :
Solution. We have
sþ2
ðs2 þ 4s þ 13Þ
Therefore
(
1
L
2
sþ2
ðs2 þ 4s þ 13Þ2
¼
sþ2
ððs þ 2Þ2 þ 32 Þ
)
(
2t 1
¼e
L
2
:
s
ðs2 þ 32 Þ2
)
:
Let
s
ðs þ
32 Þ 2
¼
s
1
:
¼ FðsÞ gðsÞ:
s 2 þ 3 2 s 2 þ 32
Then f ðtÞ ¼ L1 fFðsÞg ¼ cos 3t and
sin 3t
:
gðtÞ ¼ L1 fGðsÞg ¼
3
By Convolution Theorem
Z t
sin 3ðt uÞ
1
du
cos 3u
L fFðsÞ G ðsÞg ¼ f g ¼
3
0
Z
1 1 t
¼ ½sin 3t þ sin 3ðt 2uÞdu
3 2 0
1
1
¼ sin 3tjujt0 þ ½cos 3ðt 2uÞt0
6
12
6.28
n
Engineering Mathematics-II
1
1
¼ t sin 3t þ ½cos 3ðtÞ cos 3t
6
12
1
¼ t sin 3t:
6
Hence (
1
L
sþ2
ðs2 þ 4s þ 13Þ2
)
¼e
2t
EXAMPLE 6.81. Apply Convolution Theorem to evaluate
2
1
1
L
:
s 2 þ a2
Solution. Proceeding as in Example 6.56, we have
1
:
FðsÞ ¼ GðsÞ ¼ 2
s þ a2
Then
sin at
f ðtÞ ¼ L1 fFðsÞg ¼
a
sin
at
gðtÞ ¼ L1 fGðsÞg ¼
:
a
Therefore, by Convolution Theorem,
L1 fFðsÞGðsÞg
Z t
sin au sin aðt uÞ
:
du
¼f g¼
a
a
0
Z
1 t1
¼ 2
½cos a ð2u tÞ cos at du
a 0 2
Z t
Z t
1
¼ 2
cos a ð2u tÞdu cos at du
2a
0
0
¼
Then
f ðtÞ ¼ L1 fFðsÞg ¼ sin t
1
t sin 3t
6
1
¼ t e2t sin 3t:
6
1
¼ 2
2a
Solution. Let
s
1
s
¼ 2
: 2
¼ FðsÞGðSÞ:
2
2
ðs þ 1Þðs þ 4Þ ðs þ 1Þ s þ 4
sin að2u tÞ t
t cos at
2a
0
1 sin at sin at
þ
t cos at
2a2 2a
2a
1 sin at
t cos at
2a2
a
1
¼ 3 ½sin at at cos at:
2a
¼
EXAMPLE 6.82. Find the inverse Laplace transform of
s
ðs2 þ1Þðs2 þ4Þ using Convolution Theorem.
gðtÞ ¼ L1 fGðsÞg ¼ cos 2t:
By Convolution Theorem,
L1 fFðsÞGðsÞg
¼ sin t cos 2t
Z t
¼
sin u cos 2ðt uÞdu
0
Z
1 t
½sinð2t uÞ sinð2t 3uÞdu
¼
2 0
1 cosð2t uÞ cosð2t 3uÞ t
þ
2
1
3
0
1
1
1
¼ cos t cos t cos 2t þ cos 2t
2
3
3
1
1
1
¼ cos t cos 2t ¼ ðcos t cos 2tÞ:
3
3
3
¼
EXAMPLE 6.83. Using Convolution Theorem, find
s
and
inverse Laplace transform of (a) ðs2 þa
2 Þ3
1
(b) sðsþ1Þðsþ2Þ
Solution. (a) Let
FðsÞ ¼
GðsÞ ¼
s
and
s 2 þ a2
1
ðs2 þ a2 Þ2
Then (see Example 6.59)
1
½sin at at cos at:
2a3
By Convolution Theorem, we have
f ðtÞ ¼ cos at and gðtÞ ¼
L1 fFðsÞGðsÞg
Z t
1
cos au½sin aðt uÞ
¼f g¼
3
2a
0
aðt uÞ cos aðt uÞdu
Z t
1
¼ 3
½cos au sin aðt uÞ
2a 0
a cos auðt uÞ cos aðt uÞdu
n
Inverse Laplace Transform
Z t
1
cos au sin aðt uÞdu
2a3 0
Z 2
ðt uÞ cos au cos aðt uÞdu
a
1
2a3
Z
1
2a2
0
t
cos au sin aðt uÞdu
Z
t
0
Ans. 14 t sin 2t
sþ1
(c) log s1
0
¼
s
ðs2 þ4Þ2
(b)
¼
1
ðt uÞ fcos at þ cos aðt 2uÞg du
2
1
t2
1
t sin at 2 cos at 3 t sin at
3
8a
4a
8a
t
¼ 3 ½sin at at cos at:
8a
¼
Ans.
0
(f)
2s3
s2 þ4sþ13
(g)
s
s4 þ4a4
Ans. 2e2t cos 3t 73 e2t sin 3t
Hint: Use partial fraction method
Ans.
1
1
and GðsÞ ¼
:
FðsÞ ¼
sðs þ 1Þ
sþ2
aðs 2a Þ
s4 þ4a4
2
(h)
Then
1
1
s sþ1
¼ 1 et
2
(i)
s2 þ6
ðs2 þ1Þ ðs2 þ4Þ
(j)
s
s4 þs2 þ1
gðtÞ ¼ L fGðsÞg ¼ L
1
1
sþ2
¼e
2t
:
¼ e2t
0
t
ðe2u eu Þdu
0
¼ e2t
¼e
¼
2t
2u
eu
2
sin at sinh at
Ans. f (t) ¼ 13 ð5 sin t sin 2tÞ
(partial fraction method)
Ans. p2ffiffi3 sinh 2t sin
Therefore, by Convolution Theorem,
Z t
ð1 eu Þe2ðtuÞ du
L1 fFðsÞGðsÞg ¼ f g ¼
Z
1
2a2
Ans. cos at sinh at
and
1
2 sinh t
t
(d) cot1 s Hint: dsd cot1 s ¼ s2 þ
2 implies
n
o
L1 s2
þ2 ¼ tf ðtÞ implies
Ans. f(t) ¼ sint t
1 sin t ¼ tf ðtÞ
(e) 1s log 1 þ s12
Hint: Use Example 6.22
and Theorem 6.9
Rt
u
du
Ans. 2 1cos
u
(b) Let
f ðtÞ ¼ L1 fFðsÞg ¼ L1
6.29
0
1
1
et þ e2t :
2
2
EXERCISES
1. Find inverse Laplace transform of
2sþ6
Ans. te3t sin t
(a) ðs2 þ6sþ10Þ
2
t
2s2 þ5s4
s3 þs2 2s
Hint: Has simple poles, so use
residue method
(k)
Ans. 2 þ et e2t
Ans. 16 et 43 e2t þ 72 e3t
n
o
1
2. Use first shift property to find L1 pffiffiffiffiffiffi
.
sþa
n
o
n o
1
1
t21
¼ eat L1 p1ffis ¼ eat : ð1=2Þ
Hint: L1 pffiffiffiffiffiffi
sþa
(l)
2s2 4
ðsþ1Þ ðs2Þ ðs3Þ
t
e2t
1
et þ
2
2
pffiffi
3
2
eat
ffiffiffi
Ans. p
t
3. Solve Exercise 1(k) using Heaviside’s expansion formula.
4. UsenHeaviside’s
o expansion formula to find
L1
2712s
ðsþ4Þ ðs2 þ9Þ
Ans. 3e4t 3 cos 3t
pffi
5. Use series method to find L1 fe s g.
1
Ans. 2pffiffi1 t3=2 e4t
6. Show that
t2
t4
t6
ð6!Þ
L1 1s cos 1s ¼ 1 ð2!Þ
2 þ
2 þ ...
ð4!Þ2
6.30
n
Engineering Mathematics-II
7. Evaluate sint
t2.
Ans. t2 þ 2cos t 2
8. Find L{sin t
t2}.
Ans.
2
ðs2 þ1Þ s3
9. Use Convolution Theorem to find the inverse
Laplace transforms of the following :
(a)
(b)
(c)
(d)
1
sðsaÞ
2
a
ðs2 þa2 Þ2
4
s3 þs2 þsþ1
sþ2
ðs2 þ4sþ5Þ2
Ans.
eat 1
a .
Hint: Simple poles at 0 and a, satisfies growth
restriction condition, Res(0) ¼ 1/a, Res(a) ¼
eat/a
Ans. f (t) ¼ 1a(eat 1)
11. Using complex inversion formula, find the
inverse Laplace transform of the following:
(a)
s
s2 þa2
Ans. 2ðet cos t þ sin tÞ
(b)
Ans. 12 te2t sin t
(c)
1
ðsþ1Þ ðs2Þ2
1
ðs2 þ1Þ2
Ans.
1
2a ðsin at
at cos atÞ
10. Verify complex inversion formula for F(s) ¼
1
sðsaÞ.
Ans. cos at
n
12. Find L1 e
affi
.
erfc 2p
t
pffi o
a s
s
Ans.
1 t
9e
þ
1 2t
3 te
19 e2t
Ans. 12 ðsin t t cos tÞ
affi
, a > 0. Ans. 1 erf 2p
or
t
7
Applications of Laplace Transform
Laplace transform is utilized as a tool for solving
linear differential equations, integral equations,
and partial differential equations. It is also used to
evaluate the integrals. The aim of this chapter is to
discuss these applications.
7.1
ORDINARY DIFFERENTIAL EQUATIONS
Recall that a differential equation is an equation
where the unknown is in the form of a derivative.
The order of an ordinary differential equation is the
highest derivative attained by the unknown. Thus
the equation
d2y
dy
þ a þ by ¼ f ðtÞ
2
dt
dy
is of second order, whereas the equation.
3
dy
þy ¼ sinx
dx
is a first order differential equation.
Theorem 6.8, opens up the possibility of using
Laplace transform as a tool for solving ordinary
differential equations. Laplace transforms, being
linear, are useful only for solving linear differential
equations. Differential equations containing powers
of the unknown or expression such as tan x, ex
cannot be solved using Laplace transforms.
The results
Lf f 0 ðtÞg ¼ sFðsÞ f ð0Þ
and
Lf f 00 ðtÞ ¼ s2 FðsÞ sf ð0Þ f 0 ð0Þ
will be used frequently for solving ordinary differential equations. To solve linear ordinary differential equation by the Laplace transform method, we
first convert the equation in the unknown function
f (t) into an equation in F(s) and find F(s). The
inversion of F(s) then yields f (t).
Since f (0), f0 (0), and f 00 (0) appear in Laplace
transform of derivatives of f, the Laplace transform
method is best suited to initial value problems
(where auxiliary conditions are all imposed at
t ¼ 0). The solution by Laplace method with initial
conditions automatically built into it. We need not
add particular integral to complementary function
and then apply the auxiliary conditions.
(a) Ordinary Differential Equations with
Constant Solution
In case of an ordinary differential equation with
constant coefficients, the transformed equation for
F(s) turns out to be an algebraic one and, therefore, the Laplace transform method is powerful tool
for solving this type of ordinary differential equations. If
dny
d n1 y
an n þ an1 n1 þ . . . þ a0 y ¼ f ðtÞ
dt
dt
with y(0) ¼ y0, y0 (0) ¼ y1, . . ., y(n–1) (0) ¼ yn–1, then
f (t) is called input, excitation, or forcing function
and y(t) is called the output or response. Further, the
following results suggests that if f (t) is continuous
and of exponential order, then y(t) is also continuous and of exponential order.
Theorem 7.1. If an y(n) þ an–1 y(n–1) þ . . . þa0y ¼ f (t)
is nth order linear non-homogeneous equation with
constant coefficients and f is continuous on [0, 1)
and of exponential order, then y(t) is also continuous and of exponential order.
EXAMPLE 7.1
Find the general solution of the differential equation
y00 ðtÞ þ k 2 yðtÞ ¼ 0:
Solution. Assume that the value of the unknown
function at t ¼ 0 be denoted by the constant A, and
7.2
n
Engineering Mathematics-II
the value of its first derivative at t ¼ 0 by the
constant B. Thus
yð0Þ ¼ A and y0 ð0Þ ¼ B:
Taking Laplace transform of both sides of the given
differential equations, we have
Lfy00 ðtÞg þ k 2 LfyðtÞg ¼ 0
But
Lfy00 ðtÞg ¼ s2 YðsÞ syð0Þ y0 ð0Þ
¼ s2 YðsÞ As B:
We note that x(t) ! 1 as t ! 1 due to the term
t cos t. This term is called a secular term. The
presence of secular term causes resonance, because
the solution becomes unbounded.
Remark 7.1. If we consider the equation
þ k2x ¼ A sin t, k 6¼ 1, then there will be no
secular term in the solution and so the system will
be purely oscillatory.
d2 x
dt2
EXAMPLE 7.3
Solve the initial value problem
Therefore,
y0 ðtÞ þ 3yðtÞ ¼ 0; yð0Þ ¼ 1:
s2 YðsÞ As B þ k 2 YðsÞ ¼ 0:
Solution. Taking Laplace transform, we get
The solution of this algebraic equation in Y(s) is
s
B
k
YðsÞ ¼ A 2
þ : 2
:
2
s þk
k s þ k2
Taking inverse Laplace transform, we get
B
sin kt;
k
where A and B are constants since the initial conditions were not given.
yðtÞ ¼ A cos kt þ
Lfy0 ðtÞg þ 3LfyðtÞg ¼ 0;
which yields
sYðsÞ yð0Þ þ 3YðsÞ ¼ 0:
Since y(0) ¼ 1, we have
sYðsÞ 3YðsÞ ¼ 1;
an algebraic equation whose solution is
YðsÞ ¼
EXAMPLE 7.2
Solve
1
:
sþ3
Taking inverse Laplace transform leads to
yðtÞ ¼ e3t :
2
d x
þ x ¼ A sin t; xð0Þ ¼ x0 ; x0 ð0Þ ¼ v0 :
dt2
Show that the phenomenon of resonance occurs in
this case.
Solution. Taking Laplace transform, we get
s2 XðsÞ sxð0Þ x0 ð0Þ þ XðsÞ ¼
A
s2 þ 1
EXAMPLE 7.4
Solve the initial value problem
d2y
dy
2 8y ¼ 0; yð0Þ ¼ 3;
2
dt
dt
y0 ð0Þ ¼ 6:
Solution. The given equation is
or
ðs2 þ 1Þ XðsÞ ¼
A
þ sx0 þ v0
2
s þ1
y00 ðtÞ 2y0 ðtÞ 8y ¼ 0; yð0Þ ¼ 3; y0 ð0Þ ¼ 6:
Laplace transform leads to
or
Lfy00 ðtg 2Lfy0 ðtÞg 8Lfyg ¼ 0;
A
s
v0
x0 þ 2
:
þ 2
XðsÞ ¼
2
2
s þ1
s þ1
ðs þ 1Þ
Taking inverse Laplace transform, we have
xðtÞ ¼
A
ðsin t t cos tÞ þ x0 cos t þ v0 sin t:
2
that is,
s2 YðsÞsyð0Þy0 ð0Þ2fsYðsÞyð0Þg8YðsÞ¼0
and so using initial conditions, we have
ðs2 2s 8ÞYðsÞ 3s ¼ 0:
Applications of Laplace Transform
Hence
"
#
3s
s1þ1
¼3
2
s 2s 8
ðs 1Þ2 9
"
#
s1
1
þ
:
¼3
ðs 1Þ2 9 ðs 1Þ2 9
YðsÞ ¼
Taking inverse Laplace transform, we get
(
)
(
s1
1
1
1
yðtÞ ¼ 3L
þ 3L
ðs 1Þ2 9
ðs 1Þ2 9
n
7.3
Verification: We have
1
1
1
y0 ¼ 1 þ t2 þ et þ et ;
2
2
2
1
1
y00 ¼ t et þ et ;
2
2
1
1
y000 ¼ 1 þ et þ et :
2
2
00
000
and
y
,
we
get
Adding
y
)
y00 þ y000 ¼ t þ et þ 1 (the given equationÞ:
¼ 3et cosh 3t þ et sinh 3t:
EXAMPLE 7.6
Solve
d2y
dy
þ 2 3y ¼ sin t; yð0Þ ¼ y0 ð0Þ ¼ 0:
2
dt
dt
EXAMPLE 7.5
Solve the initial value problem
y000 þ y00 ¼ et þ t þ 1; yð0Þ ¼ y0 ð0Þ ¼ y00 ð0Þ ¼ 0:
Solution. Taking Laplace transform of both sides of
the given equation, we take
Solution. Taking Laplace transform of both sides of
the given equation, we have
Lfy000 ðtÞg þ 2Lfy0 ðtÞg 3LfyðtÞg ¼ Lfsin tg;
Lfy000 ðtÞ þ Lfy00 ðtÞg ¼ Lfet g þ Lftg þ Lf1g;
that is,
s3 YðsÞ s2 yð0Þ sy0 ð0Þ y00 ð0Þ
þ s2 YðsÞ syð0Þ y0 ð0Þ ¼
1
1 1
þ þ
s 1 s2 s
which yields
s2 YðsÞ syð0Þ y0 ð0Þ þ 2fsYðsÞ yð0Þg 3YðsÞ
1
:
¼ 2
s þ1
Using the given initial conditions, we have
s2 YðsÞ þ 2sYðsÞ 3YðsÞ ¼
Since y(0) ¼ y0 (0) ¼ y 00 (0) ¼ 0, we have
s3 YðsÞ þ s2 YðsÞ ¼
1
1 1
þ 2þ ;
s1 s
s
and so
s1
sþ1
2
2
2ðs þ1Þ 2ðs þ2s3Þ
"
#
s
1
1
sþ1
¼ 2
:
2ðs þ1Þ 2ðs2 þ1Þ 2 ðsþ1Þ2 4
YðsÞ¼
and so
YðsÞ ¼
2s2 1
:
s4 ðs 1Þ ðs þ 1Þ
Using partial fraction decomposition, we have
1
1
1
1
YðsÞ ¼ 2 þ 4 :
s
s
2ðs þ 1Þ 2ðs 1Þ
1
s2 þ 1
1
ðs2 þ1Þðs2 þ2s3Þ
¼
Taking inverse Laplace transform, we have
1
1
1
yðtÞ ¼ cos t sin t et sinh 2t:
2
2
2
Taking inverse transform yields
1
1
1
1
þ þ
s2 s4 2ðs þ 1Þ 2ðs 1Þ
1
1
1
¼ t þ t3 et þ et :
6
2
2
yðtÞ ¼ L1 EXAMPLE 7.7
Solve
d2y
dy
6 þ 9y ¼ t2 e3t ; yð0Þ ¼ 2; y0 ð0Þ ¼ 6:
dt2
dt
7.4
n
Engineering Mathematics-II
Solution. Taking Laplace transform, we get
0
s YðsÞ syð0Þ y ð0Þ 6ðsYðsÞ yð0ÞÞ þ 9YðsÞ
2
¼
:
ðs 3Þ3
2
Using initial conditions, we have
s2 YðsÞ 2s 6 6sYðsÞ þ 12 þ 9YðsÞ
¼
Solution. Taking Laplace transform leads to
1 es
sYðsÞ þ 2 þ 2YðsÞ ¼ s
s
or
ðs þ 2ÞYðsÞ ¼
or
YðsÞ ¼
2
ðs 3Þ3
1 es
2
s
s
1
es
2
:
sðs þ 2Þ sðs þ 2Þ s þ 2
But, by partial fraction, we have
or
ðs2 6s þ 9ÞYðsÞ ¼ 2ðs 3Þ þ
1
1
1
¼ :
sðs þ 2Þ 2s 2ðs þ 2Þ
2
ðs 3Þ3
Therefore,
or
2
2
:
þ
YðsÞ ¼
s 3 ðs 3Þ5
Taking inverse Laplace transform yields
1
yðtÞ ¼ 2e3t þ t4 e3t :
12
EXAMPLE 7.8
Solve
y00 3y0 þ 2y ¼ t;
yð0Þ ¼ 0 and y0 ð0Þ ¼ 0:
Solution. Taking Laplace transform yields
s2 YðsÞ syð0Þ y0 ð0Þ 3½sYðsÞ yð0Þ
1
þ 2YðsÞ ¼ 2 :
s
Making use of initial value conditions, we have
1
ðs2 3s þ 2ÞYðsÞ ¼ 2
s
and so
1
3s þ 2Þ
1
1
3
1
þ þ
¼
:
4ðs 2Þ ðs 1Þ 4s 2s2
YðsÞ ¼
s2 ðs2
Taking inverse Laplace transform, we get
1
3 t
yðtÞ ¼ e2t et þ þ :
4
4 2
Taking inverse transform, we get
1 1
1
1
yðtÞ¼ e2t Hðt1Þþ e2ðt1Þ Hðt1Þþ2e2t
2 2
2
2
(
1 3 2t
þ
e
for
0t<1
2 2
¼ 3 2t
1 2ðt1Þ
þ2e
for t 1:
2e
EXAMPLE 7.10
Solve
d2y
dy
þ2 þ5y ¼ et sint; yð0Þ ¼ 0; y0 ð0Þ ¼ 1:
2
dt
dt
Solution. Taking Laplace transform yields
s2 YðsÞ syð0Þ y0 ð0Þ þ 2ðsYðsÞ yð0ÞÞ
1
þ 5YðsÞ ¼
:
ðs þ 1Þ2 þ 1
Using initial conditions, we get
s2 YðsÞ þ 2sYðsÞ þ 5YðsÞ ¼ 1 þ
1
ðs þ 1Þ2 þ 1
or
ðs2 þ 2s þ 5ÞYðsÞ ¼ 1 þ
1
ðs þ 1Þ2 þ 1
or
EXAMPLE 7.9
Solve
y0 þ 2y ¼ 1 Hðt 1Þ;
s
1
1
e
es
2
YðsÞ ¼ :
2s 2ðs þ 2Þ
2s 2ðs þ 2Þ
sþ2
1
1
þ
s2 þ 2s þ 5 ðs2 þ 2s þ 5Þðs2 þ 2s þ 2Þ
s2 þ 2s þ 3
:
¼ 2
ðs þ 2s þ 5Þ ðs2 þ 2s þ 2Þ
Y ðsÞ ¼
yð0Þ ¼ 2;
where H(t) is Heaviside’s unit step function.
Applications of Laplace Transform
Using partial fractions, we have
Hence
2
1
þ
YðsÞ ¼
3ðs2 þ 2s þ 5Þ 3ðs2 þ 2s þ 2Þ
2
¼
:
3ððs þ 1Þ2 þ 1Þ
Taking inverse transform, we get
2 1
1
yðtÞ ¼ : et sin 2t þ et sin t
3 2
3
1 t
¼ e ðsin t þ sin 2tÞ:
3
XðsÞ ¼
n
7.5
mv0
mv0
:
ks
kðs þ mk Þ
Then, application of inverse Laplace transform yields
mv0
k
ð1 emt Þ:
xðtÞ ¼
k
The graph of x(t) is shown in the Figure 7.1
x(t )
mv0 / k
EXAMPLE 7.11
Solve the equation of motion
d2x
dx
þ k ¼ mv0 ðtÞ; xð0Þ ¼ x0 ð0Þ ¼ 0;
dt2
dt
which represents the motion of a pellet of mass m
fired into a viscous gas from a gun at time t ¼ 0 with
a muzzle velocity v0 and where (t) is Dirac delta
function, x(t) is displacement at time t 0 and k > 0
is a constant.
m
0
Solution. The condition x (0) ¼ 0 implies that the
pellet is initially at rest for t < 0. Taking the Laplace
transform of both sides, we have
t
0
Figure 7.1
The velocity is given by
dx
k
¼ x0 ðtÞ ¼ v0 em t :
dt
m½s2 XðsÞ sxð0Þ x0 ð0Þ þ k½s XðsÞ xð0Þ ¼ 1:mv0 : We observe that lim x0 (t) ¼ v0 and lim x0 (t) ¼ 0.
t!0þ
t!0
Using the given conditions, this expression reduces to This indicates instantaneous jump in velocity at
ðms2 þ ksÞ XðsÞ ¼ mv0
t ¼ 0 from a rest state to the value v0. The graph of
or
x0 (t) is shown in the Figure 7.2.
mv0
v0
¼ XðsÞ ¼ 2
ms þ ks s s þ mk
x ′(t )
Use of partial fractions yields
or
v0
A
B
¼ þ
XðsÞ ¼ s s þ mk
s s þ mk
v0
k
v0 ¼ A s þ
þ Bs:
m
Comparing coefficients, we get
v0 ¼ A
k
mv0
; which yields A ¼
k
m
and
0 ¼ A + B, which gives B ¼ mv0
:
k
t
0
Figure 7.2
7.6
n
Engineering Mathematics-II
EXAMPLE 7.12
Solve boundary value problem
d2y
þ 9y ¼ cos 2t; yð0Þ ¼ 1;
dt2
Comparing coefficients of the powers of s, we get
Solution. Suppose that y0 (0) ¼ A. Then taking
Laplace transform, we have
s
s2 YðsÞ syð0Þ y0 ð0Þ þ 9 YðsÞ ¼ 2
s þ4
or
ðs2 þ 9ÞYðsÞ ¼ s þ A þ
and so
3
;
50
2
D¼ :
25
A¼
y
¼ 1:
2
sþA
s
þ 2
2
s þ 9 ðs þ 9Þ ðs2 þ 4Þ
4s
A
s
þ
þ
¼
5ðs2 þ 9Þ s2 þ 9 5ðs2 þ 4Þ
(partial fractionsÞ:
Taking inverse Laplace transform yields
4
A
1
yðtÞ ¼ cos 3t þ sin 3t þ cos 2t:
5
3
5
Since y 2 ¼ 1, putting t ¼ 2, we get A ¼ 12
5 . Hence
4
4
1
yðtÞ ¼ cos 3t þ sin 3t þ cos 2t:
5
5
5
EXAMPLE 7.13
Solve
d2x
dx
þ 6 þ 9x ¼ sin t ðt 0Þ
dt2
dt
subject to the conditions x(0) ¼ x0 (0) ¼ 0.
1
;
10
C¼
3
;
50
Hence
XðsÞ ¼
s
;
s2 þ 4
YðsÞ ¼
B¼
3
1
3s
þ
50ðs þ 3Þ 10ðs þ 3Þ2 50ðs2 þ 1Þ
þ
2
:
25ðs2 þ 1Þ
Application of inverse Laplace transform gives
xðtÞ ¼
¼
3 3t e3t t 3
2
e þ
cos t þ sin t
50
10
50
25
e3t
3
2
ð5t þ 3Þ cos t þ sin t:
50
50
25
3t
The term e50 (5t þ 3) is the particular solution,
called the transient response since it dies away for
3
2
cos t þ 25
sin t
large time, whereas the terms – 50
is called the complementary function (sometimes
called steady state response by engineers since
it persists). However, there is nothing steady
about it.
(b) Problems Related to Electrical Circuits
Consider the RCL circuit, shown in the Figure 7.3,
consisting of resistance, capacitor, and inductor
connected to a battery.
Solution. Taking Laplace transform of both sides of
the given equations yields
L
0
s XðsÞ sxð0Þ x ð0Þ þ 6ðsXðsÞ xð0ÞÞ
1
:
þ 9XðsÞ ¼ 2
s þ1
Using the initial conditions, we have
1
ðs2 þ 6s þ 9ÞXðsÞ ¼ 2
þ1
s
or
2
XðsÞ ¼
E
R
C
Figure 7.3
1
2
ðs2 þ 1Þ ðs þ 3Þ
A
B
Cs þ D
:
þ
¼
þ 2
s þ 3 ðs þ 3Þ2
s þ1
We know that resistance R is measured in ohms,
capacitance C is measured in farads, and inductance
is measured in henrys.
Applications of Laplace Transform
Let I denote the current flowing through the circuit and Q denote the charge. Then current I is
related to Q by the relation I ¼ dQ
dt . Also
(a) By Ohm’s law, VI ¼ R (resistance). Therefore,
the voltage drop V across a resistor R is RI.
(b) The voltage drop across the inductor L is
L dI
dt .
(c) The voltage drop across a capacitor is Q
C.
Thus, if E is the voltage (potential difference) of the
battery, then by Kirchhoff’s law, we have
dI
Q
L þ RI þ ¼ EðtÞ;
dt
C
where L, C, and R are constants. In terms of current,
this equation becomes
Zt
dI
1
IðuÞ du ¼ EðtÞ;
L þ RI þ
dt
C
0
because I ¼
dQ
dt
implies Q ¼
Rt
L
L
d Q
dQ Q
þ ¼ EðtÞ;
þR
dt2
dt C
EXAMPLE 7.14
Given that I ¼ Q ¼ 0 at t ¼ 0, find I in the LR circuit
(Figure 7.4) for t > 0.
E0 v
;
s2 þ v 2
where F(s) denotes the Laplace transform of I.
Using the given initial condition, we have
ðLs þ RÞFðsÞ ¼
E0 v
s2 þ v 2
which yields
E0 vL
E0 v
¼
ðLs þ RÞ ðs2 þ v2 Þ
s þ RL ðs2 þ v2 Þ
A
Bs þ C
þ 2
:
¼
R
s þ v2
sþL
FðsÞ ¼
Comparison of coefficients of different powers of
s yields
E0 Lv
E0 Lv
E0 Rv
; B¼ 2 2
; C¼ 2 2
:
L2 v2 þ R2
L v þ R2
L v þ R2
Hence
E Lv
sE0 Lv
0
sþ RL ðL2 v2 þR2 Þ ðs2 þv2 Þ ðL2 v2 þR2 Þ
E0 Rv
:
þ 2
2
ðs þv Þ ðL2 v2 þR2 Þ
FðsÞ ¼ Taking inverse Laplace transform yields
IðtÞ ¼
L
Ið0Þ ¼ 0;
L½sFðsÞ Ið0Þ þ RFðsÞ ¼
A¼
which is a differential equation of second order with
constant coefficients L, R, and 1/C. The forcing
function (input function) E(t) is supplied by the
battery (voltage source). The system described by
the above differential equation is known as harmonic oscillator.
dI
þ RI ¼ E0 sin vt;
dt
where L, R, E0, and v are constants. Taking Laplace
transform of both sides, we have
IðuÞ du.
2
7.7
Solution. By Kirchhoff’s law, the differential equation governing the given circuit is
0
In terms of charge, this differential equation
takes the form
n
E0 Lv
E0 Lv
R
e L t 2 2
cos vt
L2 v2 þ R2
L v þ R2
E0 R
sin vt:
þ 2 2
L v þ R2
E 0 sin ωt
R
Figure 7.4
EXAMPLE 7.15
Given that I ¼ Q ¼ 0 at t ¼ 0, find charge Q and
current I in the following circuit (Figure 7.5) for
t > 0.
7.8
n
Engineering Mathematics-II
Suppose the roots of s2 þ
Then
1 henry
s1 ¼
1sin t
6 ohms
1/9 farad
Figure 7.5
Solution. By Kirchhoff’s law, the differential equation for the given circuit is
L
d2Q
dQ Q
þ ¼ EðtÞ:
þR
dt2
dt C
Here L ¼ 1, R ¼ 6, C ¼ 19, E(t) ¼ sin t. Thus we have
d2Q
dQ
þ6
þ 9Q ¼ sin tðt > 0Þ;
2
dt
dt
subject to Q(0) ¼ 0, Q0 (0) ¼ I(0) ¼ 0. By Example
7.13, the solution of this equation is
Rþ
R
Ls
1
þ LC
are s1 and s2.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
R2 ð4L=CÞ
R R2 ð4L=CÞ
and s2 ¼
:
2L
2L
Let us suppose R > 0. Then three cases arise:
(a) If R2 4L
C < 0, then s1 and s2 are complex and
s1 ¼ s2 .
(b) If R2 4L
C ¼ 0, then s1 and s2 are real and s1 ¼ s2.
2
4L
(c) If R C > 0, then s1 and s2 are real and s1 6¼ s2.
Case (a). Using partial fractions, we have
1
1
1
1
¼
:
QðsÞ¼
Lðss1 Þ ðss2 Þ Lðs1 s2 Þ ss1 ss2
Taking inverse Laplace transform yields
1
qðtÞ¼
:
Lðs1 s2 Þ½es1 t es2 t If we put
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1 4L
R
v0 ¼
R2 and ¼ ;
2L C
L
then s1 ¼ s2 ¼ þ iv0 and so s1 – s2 ¼ 2iv0.
Then
Therefore,
dQ 5e3t 3
3
2
1 ðþiv0 Þt
ð5t þ3Þe3t þ sint þ cost
IðtÞ ¼ ¼
e
eðiv0 Þt
qðtÞ ¼
50 50
dt
50
25
2Liv0
e3t
3
2
1 t eiv0 t eiv0 t
¼
ð15t þ4Þþ sint þ cost:
¼
e
50
50
25
2i
Lv0
1
t
e sin v0 t; < 0:
¼
EXAMPLE 7.16
Lv0
Solve
Thus, the impulse response q(t) is a damped sinusoidal with frequency v0. That is why, this case is
2
d q
dq q
called damped vibration or undercritical damping
L 2 þ R þ ¼ ðtÞ (Dirac delta functionÞ
dt
dt C
(Figure 7.6).
under conditions q(0) ¼ q0 (0) ¼ 0.
e3t
3
2
ð5t þ 3Þ cos t þ sin t:
QðtÞ ¼
50
50
25
1
Lω0
Solution. Applying Laplace transform to both sides
of the given equation, we find
1
Ls2 þ Rs þ
QðsÞ ¼ 1
C
2 π /ω0
π /ω 0
or
QðsÞ ¼
Ls2
1
1
:
¼ 1
þ Rs þ C1 L s2 þ RL s þ LC
−
1
L ω0
Figure 7.6
t
Applications of Laplace Transform
Case (b) In this case s1 ¼ s2 ¼ –
QðsÞ ¼
R
2L
1
Lðs Þ2
n
7.9
and so
:
Taking inverse transform, we get
qðtÞ ¼
tet
;
L
< 0:
t
0
This case is called critical damping (Figure 7.7)
Figure 7.8
(c) Mechanical System (Mass-Spring System)
− 1/ σ
0
t
− 1/ σ
Figure 7.7
Case (c) As in case (a), we have
1
ðes1 t es2 t Þ:
Lðs1 s2 Þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Since L > 0 and C > 0, we have R > R2 4L
C and
so s2 < s1 < 0. Thus q(t) is the sum of two exponentially damped functions. Put
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
4L
R
v0 ¼
R2 and ¼ :
2L
C
2L
qðtÞ ¼
Then, we have
Let m be the mass suspended on a spring that is
rigidly supported from one end (Figure 7.9). The
rest position is denoted by x ¼ 0, downward displacement by x > 0, and upward displacement is
represented by x < 0. Let
(i) k > 0 be the spring constant (or stiffness)
and a > 0 be the damping constant.
(ii) a dx
dt be the damping force due to medium
(air, etc.). Thus, damping force is proportional to the velocity.
(iii) f (t) represents all external impressed forces
on m. It is also called forcing or excitation.
x <0
0
s1 ¼ þ v 0 ;
m
x = 0(rest position)
x >0
f(t)
s2 ¼ v 0 :
Figure 7.9
Therefore,
s1 s2 ¼ 2v0
and
1
1
ðeðþv0 Þt eðv0 Þt Þ ¼
sinh v0 t
qðtÞ ¼
2Lv0
Lv0
1 ðþv0 Þt
e
ð1 e2v0 t Þ; < 0:
¼
2Lvo
Since þ v0 < 0, the impulse response q(t) is
damped hyperbolic sine. This case is called overdamped or overcritical damping (Figure 7.8).
By Newton’s second law of motion, the sum of
2
forces acting on m equals m ddt2x and so
m
d2x
dx
¼ kx a þ f ðtÞ:
dt2
dt
Thus the equation of motion is
d2x
dx
þ a þ kx ¼ f ðtÞ
ð1Þ
2
dt
dt
This is exactly the same differential equation which
occurs in harmonic oscillator.
m
7.10
n
Engineering Mathematics-II
If a ¼ 0, the motion is called undamped whereas
if a6¼ 0, the motion is called damped. Moreover, if
f (t) ¼ 0, that is, if there is no impressed forces, then
the motion is called forced.
The equation (1) can be written as
d 2 x a dx k
þ
¼ f ðtÞ=m;
ð2Þ
þ
dt2 m dt m
where f (t)/m is now the external impressed force
(or excitation force) per unit mass.
EXAMPLE 7.17
Solve the equation of motion
1
LfðtÞg:
þ 2bs þ l2
Thus we conclude that
XðsÞ ¼
s2
Response ¼ Transfer function Input:
(d) Ordinary Differential Equations with
Polynomial (Variable) Coefficients
We know that
dn
FðsÞ;
dsn
where F(s) ¼ L { f (t)}. Thus for n ¼ 1, we have
Lftn f ðtÞg ¼ ð1Þn
d2x
dx
þ 2b þ l2 x ¼ ðtÞ; xð0Þ ¼ x0 ð0Þ ¼ 0
2
dt
dt
for 0 < b < l.
[Clearly this is equation (2) with ma ¼ 2b, mk ¼ l2 ]
Solution. We want to find the response of the given
mechanical system to a unit impulse. Taking
Laplace transform, we get
fs2 XðsÞsxð0Þx0 ð0Þgþ2bfsXðsÞxð0Þgþl2 XðsÞ¼1:
Taking note of the given conditions, we have
ðs2 þ 2bs þ l2 Þ XðsÞ ¼ 1
or
Also we note that
1
1
¼
:
s2 þ 2bs þ l2 ðs þ bÞ2 þ ðl2 b2 Þ
Taking inverse Laplace transform yields
!
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
1
2
bt
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin l b2 t ;
xðtÞ ¼ e
l2 b2
which is clearly a case of damped oscillation
(Figure 7.10).
XðsÞ ¼
LftfðtÞ ¼ F0 ðsÞ:
Hence, if f0 (t) satisfies the sufficient condition for
the existence of Laplace transform, then
d
d
Lftf 0 ðtÞ ¼ Lf f 0 ðtÞg ¼ ðsFðsÞ f ð0ÞÞ
ds
ds
¼ sF0 ðsÞ FðsÞ:
Similarly for f 00 (t),
d
d
Lftf 00 ðtÞg¼ Lff 00 ðtÞg¼ fs2 FðsÞsf ð0Þf 0 ð0Þg
ds
ds
¼s2 F 0 ðsÞ2sFðsÞþf ð0Þ:
The above-mentioned derivations are used to solve
linear differential equations whose coefficients are
first degree polynomials.
EXAMPLE 7.18
Solve
ty00 þ y0 þ ty ¼ 0; yð0Þ ¼ 1;
y0 ð0Þ ¼ 0:
Solution. Taking Laplace transform, we have
x(t )
Lfty00 g þ Lfy0 g þ Lftyg ¼ 0
or
d
d
Lfy00 ðtÞg þ fsYðsÞ yð0Þg fYðsÞg ¼ 0
ds
ds
or
t
0
Figure 7.10
d 2
fs YðsÞ syð0Þ y0 ð0Þg þ fsYðsÞ yð0Þg
ds
d
YðsÞ ¼ 0
ds
which on using initial conditions yields
dY ðsÞ
dYðsÞ
þ 2sY ðsÞ þ sYðsÞ ¼0
s2
ds
ds
Applications of Laplace Transform
or
or
dYðsÞ
þ sYðsÞ ¼ 0
ðs þ 1Þ
ds
2
or
dYðsÞ
s ds
¼ 0:
þ 2
YðsÞ s þ 1
Integrating, we have
1
log YðsÞ þ logðs2 þ 1Þ ¼ A ðconstantÞ
2
and so
A
YðsÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffi :
s2 þ 1
Taking inverse Laplace transform, we get
yðtÞ ¼ A J0 ðtÞ;
where J0(t) is Bessel function of order zero.
Putting t ¼ 0 and using initial condition y(0) ¼ 1,
we have
1 ¼ A J0 ð0Þ ¼ A:
n
7.11
dYðsÞ
3
4
þ
s YðsÞ ¼ 2 þ 1:
ds
s
s
The integrating factor is
R 3
s2
e ðssÞ ds ¼ s3 e 2 :
Therefore,
d
4
s2
s2
s2
½YðsÞ:s3 e 2 ¼ 2 s3 e 2 þ s3 e 2 ;
ds
s
and so integration yields
Z
Z
s2
s2
s2
YðsÞs3 e 2 ¼ 4 se 2 ds þ s3e 2 ds:
2
Putting u ¼ – s2 , we get
Z
Z
2
3 s2
u
YðsÞs e ¼ 4 e du þ 2 u eu du
2
2
2
s s2
s2
s2
2
þA
¼ 4e
e e
þ2
2
Hence the required solution is
s2
s2
¼ 2e 2 s2 e 2 þ C:
yðtÞ ¼ J0 ðtÞ:
Thus,
2 1 C s2
þ
e2 :
s3 s s3
Since Y(s) ! 0 as s ! 1, we must have C ¼ 0 and
so
2 1
YðsÞ ¼ 3 :
s
s
Taking inverse Laplace transform, we get
YðsÞ ¼
EXAMPLE 7.19
Solve
y00 þ ty0 2y ¼ 4; yð0Þ ¼ 1; y0 ð0Þ ¼ 0:
Solution. Taking Laplace transform yields
Lfy00 ðtÞg þ Lfty0 ðtÞg 2LfyðtÞg ¼ 4Lf1g
yðtÞ ¼ t2 1:
or
s2 YðsÞ syð0Þ y0 ð0Þ d
4
Lfy0 ðtÞg 2YðsÞ ¼
ds
s
or
d
s2 YðsÞ syð0Þ y0 ð0Þ ðsYðsÞ yð0ÞÞ
ds
4
2YðsÞ ¼ :
s
On using the initial values, we have
dYðsÞ
4
þ YðsÞ 2YðsÞ ¼
s2 YðsÞ þ s s
ds
s
or
sdYðsÞ
4
ðs2 3ÞYðsÞ ¼ þ s
ds
s
EXAMPLE 7.20
Solve
ty00 þ 2y0 þ ty ¼ 0;
yð0Þ ¼ 1;
yðÞ ¼ 0:
Solution. Let y0 (0) ¼ A (constant). Taking Laplace
transform of both sides, we obtain
d 2
fs YðsÞ syð0Þ y0 ð0Þg þ 2fsYðsÞ yð0Þg
ds
d
fYðsÞg ¼ 0
ds
and so
s2 Y0 ðsÞ 2sYðsÞ þ yð0Þ þ 2sYðsÞ 2yð0Þ
Y0 ðsÞ ¼ 0:
7.12
n
Engineering Mathematics-II
Using boundary conditions, we get
ðs þ 1ÞY ðsÞ 1 ¼ 0
or
Y0 ðsÞ ¼
1
:
þ1
s2
Integration yields
YðsÞ ¼ tan1 s þ B (constant):
Since Y(s) tends to zero as s ! 1, we must have
B ¼ /2. Hence,
1
:
YðsÞ ¼ tan1 s ¼ tan1
2
s
Taking inverse Laplace transform, we have (see
Example 21.48).
sin t
1
1 1
:
tan
yðtÞ ¼ L
¼
s
t
This solution clearly satisfies y() ¼ 0.
EXAMPLE 7.21
Solve
00
0
ty þ y þ 2y ¼ 0;
yð0Þ ¼ 1:
Solution. Taking Laplace transform gives
d 2
ðs YðsÞ syð0Þ y0 ð0ÞÞ þ ðsYðsÞ yð0ÞÞ
ds
þ 2YðsÞ ¼ 0
or
s2 Y0 ðsÞ2sYðsÞþyð0ÞþsYðsÞyð0Þþ2YðsÞ¼0
or
s2 Y0 ðsÞ sYðsÞ þ 2YðsÞ ¼ 0
or
Y0 ðsÞ þ
or
A e s
:
s
1 n
P
x
2
Since ex ¼
n !; taking x ¼ s ; we have
n¼0
1
X
ð1Þn 2n
:
YðsÞ ¼ A
n ! snþ1
n¼0
2
0
2
1 2
2 YðsÞ ¼ 0:
s s
The integrating factor is
R 1 2 2
2
ds
¼ elog sþ s ¼ se s :
e s s2
Therefore,
d
fYðsÞse2=s g ¼ 0:
ds
Integrating, we have
YðsÞ se2=s ¼ A ðconstantÞ
YðsÞ ¼
Taking inverse Laplace transform,we get
1
X
ð1Þn 2n tn
:
yðtÞ ¼ A
ðn !Þ2
n¼0
The condition y(0) ¼ 1 now yields A ¼ 1. Hence
1
X
pffiffiffiffi
ð1Þn 2n tn
¼ J0 ð2 2tÞ;
yðtÞ ¼
2
ðn !Þ
n¼0
where J0 is Bessel’s function of order zero.
EXAMPLE 7.22
Solve
ty00 y0 ¼ 1;
yð0Þ ¼ 0:
Solution. Taking Laplace transform of both sides of
the given equation,
d
1
fs2 Y0 ðsÞsyð0Þy0 ð0ÞgfsYðsÞyð0Þg¼
ds
s
or
1
s2 Y0 ðsÞ 2sYðsÞ þ yð0Þ sY ðsÞ þ yð0Þ ¼ s
or
1
s2 Y0 ðsÞ 3sYðsÞ ¼ s
or
3
1
Y0 ðsÞ þ YðsÞ ¼ 3 :
s
s
The integrating factor is
R3
ds
e s ¼ e3 log s ¼ s3 :
Therefore,
d
1
ðYðsÞs3 Þ ¼ 3 s3 ¼ 1:
ds
s
Integrating
YðsÞs3 ¼ s þ AðconstantÞ
and so
YðsÞ ¼
1 A
þ :
s2 s3
Applications of Laplace Transform
Taking inverse Laplace transform, we get
yðtÞ ¼ t þ Bt2 ;
where B is constant. Obviously, the solution satisfies y(0) ¼ 0.
n
7.13
By initial value theorem yð0Þ ¼ lim sYðsÞ ¼ 0 and
s!0
so C ¼ 0. Hence
1
:
YðsÞ ¼
ðs þ 1Þ2
Taking inverse Laplace transform, we get
yðtÞ ¼ t et :
EXAMPLE 7.23
Solve
ty00 þ ðt þ 1Þy0 þ 2y ¼ et ; yð0Þ ¼ 0:
7.2
SIMULTANEOUS DIFFERENTIAL EQUATIONS
The Laplace transforms convert a pair of differential
equations into simultaneous algebraic equations in
parameters. After that we solve these equations for
Laplace transforms of the variables and then apply
d
d
fs2 YðsÞ syð0Þ y0 ð0Þg fsYðsÞ yð0Þg inverse Laplace operators to get the required
ds
ds
solution.
1
þ fsYðsÞ yð0Þg þ 2YðsÞ ¼
sþ1
EXAMPLE 7.24
or
Solve the simultaneous differential equations
s2 Y0 ðsÞ 2sYðsÞ fsY0 ðsÞ þ YðsÞg þ sYðsÞ
3x0 þ y0 þ 2x ¼ 1; x0 þ 4y0 þ 3y ¼ 0
1
yð0Þ þ 2YðsÞ ¼
subject to the conditions x(0) ¼ 0, y(0) ¼ 0.
sþ1
or
1
Solution. Taking Laplace transform, we get
s2 Y0 ðsÞsY0 ðsÞ2sYðsÞþsYðsÞþYðsÞ ¼
sþ1
1
3fsXðxÞ xð0Þg þ fsYðsÞ yð0Þg þ 2XðxÞ ¼
or
s
1
0
2
and
ðs þ sÞY ðsÞ ðs 1ÞYðsÞ ¼
sþ1
sXðsÞ xð0Þ þ 4fsYðsÞ yð0Þg þ 3YðsÞ ¼ 0:
or
s
1
1
Using the initial conditions, these equations reduce
¼
:
Y0 ðsÞ þ 2
s þ s sðs þ 1Þ2
to
1
ð3Þ
ð3s þ 2ÞXðsÞ þ sYðsÞ ¼
The integration factor is
s
and
R 1
R s1
2
2
ds
ds ðs þ 1Þ
s þ sþ1
:
¼e
¼
e s2 þs
sXðsÞ þ ð4s þ 3ÞYðsÞ ¼ 0:
ð4Þ
s
Multiplying (3) and (4) by s and (3s þ 2) respecTherefore,
tively and then subtracting, we get
!
d
ðs þ 1Þ2
1
ðs þ 1Þ2
1
1
1
¼
¼ 2:
:
YðsÞ
YðsÞ ¼ ¼
;
s
s
ds
s
2
sðs þ 1Þ2
11s þ 17s þ 6 ð11s þ 6Þ ðs þ 1Þ
Solution. Taking Laplace transform of both sides
gives
Integrating, we get
2
YðsÞ
ðs þ 1Þ
¼
s
Z
1
1
ds ¼ þ C
2
s
s
and so
YðsÞ ¼
1
ðs þ 1Þ2
þ
Cs
ðs þ 1Þ2
:
and then using (4), we have
4s þ 3
:
XðsÞ ¼
sð11s þ 6Þ ðs þ 1Þ
We deal with X(s) first. Using partial fraction, we
have
1
3
1
1
:
XðsÞ ¼ 2s 10 s þ ð6=11Þ
5ðs þ 1Þ
7.14
n
Engineering Mathematics-II
Taking inverse transform, we have
1 3
1
6
xðtÞ ¼ e 11t et
2 10
5
1
6
¼ ð5 3e t 2et Þ:
10
6
and –1. Hence
Further, poles of Y(s) are – 11
6
6
YðsÞ e 11t
yðtÞ ¼ lim s þ
11
s!6
11
1
6
þ lim ðs þ 1Þ YðsÞet ¼ ðet e 11 t Þ:
s! 1
5
EXAMPLE 7.25
Solve the simultaneous differential equations
dx
dy
¼ 2x 3y;
¼ y 2x
dt
dt
subject to the conditions x(0) ¼ 8, y(0) ¼ 3.
Solution. Taking Laplace transform and using the
given conditions, we have
sXðsÞ ¼ 2XðsÞ 3YðsÞ þ 8
and
Thus
and
sYðsÞ ¼ YðsÞ 2XðsÞ þ 3:
ðs 2ÞXðsÞ þ 3YðsÞ ¼ 8;
2XðsÞ þ ðs 1ÞYðsÞ ¼ 3:
Solving these algebraic equations, we get
8s 17
8s 17
XðsÞ ¼ 2
¼
;
s 3s 4 ðs þ 1Þ ðs 4Þ
and
YðsÞ ¼
3s 22
3s 22
¼
:
s2 3s 4 ðs þ 1Þ ðs 4Þ
Solution. Taking Laplace transform and using the
given conditions, we have
1
s
þ1¼
;
s1
s1
1
:
sYðsÞ þ XðsÞ ¼ 2
s þ1
sXðsÞ YðsÞ ¼
Solving these equations, we get
XðsÞ ¼
YðsÞ ¼
yðtÞ ¼ 5et 2e4t :
EXAMPLE 7.26
Solve
dx
dy
y ¼ et ;
þ x ¼ sin t
dt
dt
subject to the conditions x(0) ¼ 1, y(0) ¼ 0.
ðs 1Þ ðs2 þ 1Þ2
s4 þ s3 2s2
;
2
sðs 1Þ ðs2 þ 1Þ
¼
s3 þ s2 2s
ðs 1Þ ðs2 þ 1Þ2
:
Now
XðsÞ ¼
s4 þ s2 þ s 1
ðs 1Þ ðs2 þ 1Þ2
A
Bs þ C
Ds þ E
þ
þ
¼
:
s 1 ðs2 þ 1Þ ðs2 þ 1Þ2
Comparison of coefficients yields
1
and E ¼ 1:
A¼B¼C¼
2
Thus
1
s
1
1
þ 2
þ
þ 2
:
XðsÞ ¼
2
2ðs1Þ 2ðs þ1Þ 2ðs þ1Þ ðs þ1Þ2
Hence
1
1
1
1
xðtÞ ¼ et þ cos t þ sin t þ ðsin t t cos tÞ
2
2
2
2
1 t
¼ ½e þ cos t þ 2 sin t t cos t:
2
Now consider Y(s). We have
YðsÞ ¼
Using partial fractions, these yields
5
3
5
2
þ
; YðsÞ ¼
:
XðsÞ ¼
sþ1 s4
sþ1 s4
Hence taking inverse Laplace transform, we get
xðtÞ ¼ 5et þ 3e4t ;
s4 þ s2 þ s 1
s3 þ s2 2s
ðs 1Þ ðs2 þ 1Þ2
A
Bs þ C
Ds þ E
þ
þ
¼
:
s 1 ðs2 þ 1Þ ðs2 þ 1Þ2
Comparing coefficients, we get
1
1
1
A ¼ ; B ¼ ; C ¼ ; D ¼ 2; E ¼ 0;
2
2
2
and so
YðsÞ ¼
1
s
1
2s
þ
:
þ
2ðs1Þ 2ðs2 þ1Þ 2ðs2 þ1Þ ðs2 þ1Þ2
n
Applications of Laplace Transform
EXAMPLE 7.28
Solve the following system of equations:
Hence
1
1
1
yðtÞ ¼ et þ cos t sin t þ t sin t:
2
2
2
xðtÞ y00 ðtÞ þ yðtÞ ¼ et 1;
EXAMPLE 7.27
The co-ordinates (x, y) of a particle moving along a
plane curve at any time t are given by
dy
dx
þ 2x ¼ sin 2t;
2y ¼ cos 2t; t > 0:
dt
dt
If at t ¼ 0, x ¼ 1 and y ¼ 0, show by using transforms, that the particle moves along the curve 4x2 þ
4xy þ 5y2 ¼ 4.
Solution. Using Laplace transform, we get
sXðsÞ xð0Þ 2YðsÞ ¼
s2
x0 ðtÞ þ y0 ðtÞ yðtÞ ¼ 3et þ t;
subject to x(0) ¼ 0, y(0) ¼ 1, y0 (0) ¼ –2.
Solution. Taking Laplace transform yields
XðsÞfs2 YðsÞsyð0Þy0 ð0ÞgþYðsÞ ¼
and
sXðsÞ xð0Þ þ sYðsÞ yð0Þ YðsÞ ¼
XðsÞ s2 YðsÞ þ s 2 þ YðsÞ ¼
s
:
þ4
and
sXðsÞ þ sYðsÞ 1 YðsÞ ¼
2
¼
ðs2 þ 4Þ
s
1 2
þ
;
¼ 2
s þ 4 2 s2 þ 4
2s2 8
2
:
YðsÞ ¼
¼ 2
2
2
s þ4
ðs þ 4Þ
s2
3s2 þ s þ 1
ðs þ 1Þs2
XðsÞ ðs2 1ÞYðsÞ ¼ 2 s s
s2 þ s þ 4
¼
:
s2 þ 4
s2 þ 4
Solving for X(s) and Y(s), and using partial fractions, we have
s3 þ s2 þ 4s þ 4
1
sðs þ 1Þ
or
2
s2 þ 4
sXðsÞ 2YðsÞ ¼ 1 þ
XðsÞ ¼
3
1
þ 2:
sþ1 s
and
Using the given conditions, we have
sYðsÞ þ 2XðsÞ ¼
1
1
sþ1 s
Using the given conditions, we have
2
sYðsÞ yð0Þ þ 2XðsÞ ¼ 2
s þ4
and
7.15
¼
1
sðs þ 1Þ
s3 þ s2 þ 2s 1
sðs þ 1Þ
and
sXðsÞ þ ðs 1ÞYðsÞ ¼ 1 s
1
þ 2
þ4 s þ4
¼
3s2 s 1
ðs þ 1Þs2
s3 2s2 þ s þ 1
:
ðs þ 1Þs2
Solving for X(s) and Y(s), we have
Hence taking inverse transform, we get
1
xðtÞ ¼ cos 2t þ sin 2t;
2
yðtÞ ¼ sin 2t:
XðsÞ ¼
1
1
1
1
¼ þ
;
s2 ðs þ 1Þ s2 s þ 1 s
YðsÞ ¼
s2 s 1
1
1
¼
2:
2
s ðs þ 1Þ
sþ1 s
Hence, taking inverse Laplace transform, we get
xðtÞ ¼ t þ et 1; yðtÞ ¼ et t:
We observe that
4x2 þ 4xy þ 5y2 ¼ 4ðcos2 2t þ sin2 2tÞ ¼ 4;
and hence the particle moves along the curve 4x þ
4xy þ 5y2 ¼ 4.
2
EXAMPLE 7.29
Given that I(0) ¼ 0, find the current I in RL-network
shown in the Figure 7.11.
7.16
n
Engineering Mathematics-II
R
I
I1
L
E(t ) = 1 volt
I2
R
7.3
Figure 7.11
Solution. We note that I ¼ I1 þ I2 and so RI ¼ RI1 þ
RI2, or equivalently, RI2 ¼ RI – RI1. By Kirchhoff’s
law, we have
(a) In the closed loop containing R and L,
dI1
¼E¼1
ð5Þ
RI þ L
dt
(b) In the closed loop containing two resistances R,
RI þ RI2 ¼ E ¼ 1
or
or
Taking inverse Laplace transform yields
i
1
1 R
1 h
R
IðtÞ ¼
1 e 2L t ¼
2 e 2L t :
R
2
2R
ð6Þ
We want to solve (5) and (6) under the conditions
I(0) ¼ I1(0) ¼ 0. Taking Laplace transform yields
1
RFðsÞ þ LfsGðsÞI1 ð0Þg ¼
s
and
1
2R FðsÞ RGðsÞ ¼ :
s
Using I1(0) ¼ 0, we have
1
RFðsÞ þ LsGðsÞ ¼
ð7Þ
s
and
1
ð8Þ
2RFðsÞ RGðsÞ ¼ :
s
Multiplying (7) by R and (8) by Ls and adding, we
get
R
R þ Ls
ðR2 þ 2RLsÞFðsÞ ¼ þ L ¼
s
s
or
R þ Ls
1
R þ Ls
¼
:
FðsÞ ¼
RsðR þ 2LsÞ R sðR þ 2LsÞ
Using partial fractions, we get
1 1
1
FðsÞ ¼
:
R s 2ðs þ ðR=2LÞÞ
A relationship between the values of a function y(t)
and the values of the function at different arguments
y(t þ h), h constant, is called a difference equation.
For example,
yðn þ 2Þ yðn þ 1Þ þ yðnÞ ¼ 2
and
yðn þ 2Þ 2yðnÞ þ yðn 1Þ ¼ 1
are difference equations.
A relation between the terms of a sequence {xn}
is also a difference equation. For example,
xnþ1 þ 2xn
RI þ RI RI1 ¼ 1
2RI RI1 ¼ 1:
DIFFERENCE EQUATIONS
¼8
is a difference equation.
Difference equations (also called recurrence
relations) are closely related to differential equations and their theory is basically the same as that of
differential equations.
Order of a difference equation is the difference
between the largest and smallest arguments occurring in the difference equation divided by the unit of
increment.
For example, the order of the difference equation
¼ 2:
anþ2 3anþ1 þ 2an ¼ 5n is nþ2n
1
Solution of a difference equation is an expression
for yn which satisfies the given difference equation.
The aim of this section is to solve difference
equations using Laplace transform.
We first make the following observations:
(A) Let f (t) ¼ a[t], where [t] is the greatest integer
less than or equal to t and a > 0. Then f (t) is of
exponential order and by definition,
Z1
Lff ðtÞg ¼
e
st
Z1
f ðtÞdt¼
0
0
Z1
¼
e
0
est a½t dt
st 0
Z2
a dtþ
e
1
st 1
Z3
a dtþ
2
est a2 dtþ...
Applications of Laplace Transform
1 es aðes e2s Þ a2 ðe2s e3s Þ
þ
þ
þ ...
s
s
s
s
1e
½1 þ aes þ a2 es þ ...
¼
s
1 es
ðReðsÞ > maxð0;logaÞÞ:
¼
sð1 aes Þ
¼
(B) If L1 F{(s)} ¼ f (t), then we know that
f ðt 1Þ for t > 1
L1 fes FðsÞg ¼
0
for t < 1:
Also, by observation (1) above, we have
1 es
¼ an for n ¼ 0; 1; 2; . . . ;
L1
sð1 aes Þ
n t < n þ 1:
Therefore,
ð1 es Þes
L1
sð1 aes Þ
n t < n þ 1;
n ¼ 0; 1; 2; . . .
Then the given difference equation reduces to
yðt þ 2Þ 4yðt þ 1Þ þ 3yðtÞ ¼ 0:
Thus
Lfyðt þ 2Þg 4Lfyðt þ 1Þg þ 3LfyðtÞg ¼ 0 ð9Þ
Now
Z1
Lfzyðtþ2Þg¼
est yðtþ2Þdt
0
Z1
¼
2
esðu2Þ yðuÞdu; u¼tþ2
Z1
¼e
su
e
Z2
yðuÞdue
0
Z1
2s
esu yðuÞdu:
0
esu a0 due2s
Z2
esu a1 du
¼ fðt 1Þ
¼ an
¼ an
0
1
es e2s
since a0 ¼0;
¼e LfyðtÞge
s
a1 ¼1
es
2s
s
¼e LfyðtÞg ð1e Þ;
s
for n t 1 < n þ 1; n ¼ 0; 1; 2;
for n t < n þ 1; n ¼ 1; 2; 3; . . .
0
e
st
1
s
Z3
dtþ2a
e
st
Z4
dtþ4a
2
Hence
(
s
s
e ð1þe Þ
sð1aes Þ2
2s
2s
Z1
Lfyðtþ1Þg ¼
est yðtþ1Þdt
0
e
st
dtþ...
Z1
¼
3
e e2s
e2s e3s
e3s e4s
þ2a
þ4a2
þ...
s
s
s
es ð1es Þ
½1þ2aes þ4a2 e2s þ...
¼
s
es ð1es Þ
1
es ð1es Þ
¼
:
:
¼
s
ð1aes Þ2 sð1aes Þ2
L1
yðtÞ ¼ an ;
¼e2s LfyðtÞge2s
Z2
¼
Solution. Let us define
2s
(C) If f (t) ¼ nan–1 for n t < n þ 1, n ¼ 0, 1, 2, . . .,
then
Z1
Lf f ðtÞg¼ est f ðtÞdt
¼
7.17
n
esðu1Þ yðuÞdu; u¼tþ1
1
Z1
¼
e
sðu1Þ
Z1
yðuÞdu
0
0
Z1
¼e
s
e
su
Z1
yðuÞdue
0
¼es LfyðtÞges
¼ f ðtÞ ¼ nan1 ; n ¼ 0;1;2;...
s
esu yðuÞdu
0
Z1
)
esðu1Þ yðuÞdu
esu a0 du
0
¼es LfyðtÞg since a0 ¼0:
Hence (9) becomes
EXAMPLE 7.30
Solve
anþ2 4anþ1 þ 3an ¼ 0;
es
ð1 es Þ 4es LfyðtÞg
s
þ 3LfyðtÞg ¼ 0;
e2s LfyðtÞg a0 ¼ 0; a1 ¼ 1:
7.18
n
Engineering Mathematics-II
which yields
Then the difference equation becomes
s
yðt þ 2Þ 4yðt þ 1Þ þ 3yðtÞ ¼ 5n :
ð10Þ
e ð1 e Þ
2s
s
sðe 4e þ 3Þ
By observation (B) and Example 7.30, we have
es ð1 es Þ
1
1
1 es
n
s
¼
;
g
¼
Lf5
s
2s
e 3 e 1
sð1 5es Þ
1 es
1
1
es
2s
¼
ð1 es Þ;
Lfyðt
þ
2Þg
¼
e
LfyðtÞg
2s
1 3es 1 es
s
1 es
1 es
Lfyðt þ 1Þg ¼ es LfyðtÞg:
¼
s
s
2sð1 3e Þ 2sð1 e Þ
Taking Laplace transform of both sides of (10), we
1
1
½t
¼ Lf3 g Lf1g; by observation ð1Þ: have
2
2
Lfyðt þ 2Þg 4Lfyðt þ 1Þg þ 3LfyðtÞg ¼ Lf5½t g
Hence inversion yields
or
1 n
an ¼ ½3 1; n ¼ 0; 1; 2; . . .
es
2
e2s LfyðtÞg ð1 es Þ 4es LfyðtÞg
s
EXAMPLE 7.31
þ 3LfyðtÞg ¼ Lf5½tg
Solve the difference equation
or
s
LfyðtÞg ¼
yðt þ 1Þ yðtÞ ¼ 1; yðtÞ ¼ 0; t < 1:
Solution. Taking Laplace transformation of both
sides, we get
Lfyðt þ 1Þg LfyðtÞg
¼ Lf1g:
fe2s 4es þ 3gLfyðtÞg ¼
Hence
LfyðtÞg ¼
But, as in Example 7.30, we have
Lfyðt þ 1Þg
s
But
or
LfyðtÞg
¼
1
:
sðes 1Þ
Taking inverse Laplace transform, we have
yðtÞ
¼ L1
sðes
1
1Þ
¼ ½t; t > 0ðsee Example 20:3Þ
EXAMPLE 7.32
Solve
anþ2 4anþ1 þ 3an ¼ 5n ;
a0 ¼ 0;
a1 ¼ 1
Solution. We define
yðtÞ
¼ an ;
n t < n þ 1;
es ð1 es Þ
Lf5½t g
þ
sðe2s 4es þ 3Þ e2s 4es þ 3
1
1
Lf5½t g
¼ Lf3½t g Lf1g þ 2s
:
2
2
e 4es þ 3
¼ e LfyðtÞg;
1
es LfyðtÞg LfyðtÞg ¼
s
and so
es ð1 es Þ
þ Lf5½t g:
s
n ¼ 0; 1; 2; . . .
Lf5½t g
1es
1
¼
: 2s
2s
s
s
e 4e þ3 sð15e Þ e 4es þ3
es 1
¼ s
sðe 5Þðes 3Þðes 1Þ
es 1 1=8
1=4
1=8
þ
¼
s
es 5 es 3 es 1
1es
1=8
1=4
1=8
þ
¼
s
15es 13es 1es
1
1
1
¼ Lf1gþ Lf5½t g Lf3½t g:
8
8
4
Hence
3
1
1
LfyðtÞg ¼
Lf1g þ Lf3½t g þ Lf5½t g
8
4
8
and so
3 1 n 1 n
an ¼
þ 3 þ 5 :
8
4
8
Applications of Laplace Transform
EXAMPLE 7.33
Solve
anþ2 3anþ1 þ 2an ¼ 2n ;
n
7.19
Therefore,
a0 ¼ 0;
a1 ¼ 1:
LfyðtÞg
¼ Lfn 2n1 g:
Hence
an ¼ n 2n1 ;
Solution. We define
yðtÞ ¼ an ; n t < n þ 1:
n ¼ 0; 1; 2; . . .
Verification. We note that
anþ1 ¼ ðn þ 1Þ2n ; anþ2 ¼ ðn þ 2Þ2nþ21
Then the given equation reduces to
¼ ðn þ 2Þ2nþ1 :
yðt þ 2Þ 3yðt þ 1Þ þ 2yðtÞ ¼ 2½t :
Taking Laplace transform of both sides, we get
Therefore,
Lfyðt þ 2Þg 3Lfyðt þ 1Þg þ 2LfyðtÞg ¼ Lf2½t g:
anþ2 3anþ1 þ2an ¼ðnþ2Þ2nþ1 ð3nþ3Þ2n þ2n2n1
¼2n ½2nþ43n3þn¼2n :
But, as in the previous examples,
Lfyðt þ 2Þg ¼ e2s LfyðtÞg es
ð1 es Þ;
s
EXAMPLE 7.34
Solve
yðtÞ yðt Þ ¼ sin t;
Lfyðt þ 1Þg ¼ es LfyðtÞg:
Therefore,
t 0:
Solution. Taking Laplace transform, we have
ðe2s 3es þ 2ÞLfyðtÞg ¼
LfyðtÞg Lfyðt Þg ¼ Lfsin tg:
s
e
ð1 es Þ þ Lf2½t g;
s
e
s
sðe2s 3es þ2Þ
ð1es Þþ
½t
Lf2 g
e2s 3es þ2
ð11Þ
But
Z1
which gives
LfyðtÞg¼
yðtÞ ¼ 0;
LfyðtÞg ¼
est yðt Þdt
0
Z1
es ð1es Þ
1
1
Lf2½t g
¼
þ
esðuþÞ yðuÞdu; u ¼ t ¼
s
es 2 es 1
e2s 3es þ2
1es
1
1
Lf2½t g
þ
¼
Z1
s
12es 1es
e2s 3es þ2
s
esu yðuÞdu; yðuÞ ¼ 0; u < 0
¼e
1es
1
Lf2½t g
þ
¼
0
sð1es Þ s e2s 3es þ2
s
¼
e
LfyðtÞg:
Lf2½t g
¼Lf2½t gLf1gþ 2s
:
Hence (11) reduces to
e 3es þ2
1
Lf yðtÞg es Lf yðtÞg ¼ 2
But
s þ1
and so
Lf2½t g
1es
1
1
¼
:
Lf yðtÞg ¼ 2
:
e2s 3es þ2 sð12es Þ e2s 3es þ2
ðs þ 1Þ ð1 es Þ
es 1
Taking inverse transform, we get (see Exapmle 5.72)
¼ s
sðe 2Þðes 2Þðes 1Þ
"
#
sin t for 0 < t < yðtÞ ¼
es 1 1
1
1
0
for < t < 2
þ
þ
¼
s
es 2 ðes 2Þ2 es 1
sin t for 2n < t < ð2n þ 1Þ
"
#
¼
1es
1
es
1
0
for ð2n þ 1Þ < t < ð2n þ 2Þ
¼
þ
þ
s
12es ð12es Þ2 1es
(due to periodicity), for n¼ 0; 1; 2; . . .
¼ Lf2½t gþLfn 2n1 gþLf1g:
This is half-wave rectified sinusoidal function.
7.20
n
Engineering Mathematics-II
EXAMPLE 7.35
Find explicit formula (solution) for Fibonacci
sequence:
anþ2 ¼ anþ1 þ an ; a0 ¼ 0; a1 ¼ 1:
But
Z1
Lfyðt1Þg¼
0
Z1
Solution. Define
yðtÞ ¼ an ;
est yðt1Þdt
n t < n þ 1;
¼
n ¼ 0; 1; 2; . . .
esðuþ1Þ yðuÞdu; u¼t1
1
Then the given difference equation reduces to
yðt þ 2Þyðt þ 1ÞyðtÞ ¼ 0:
¼e
s
Taking Laplace transform, we have
¼es
But
ðe2s es 1ÞLfyðtÞg ¼
yðuÞduþe
s
Z1
Z1
esu yðuÞdu
0
esu yðuÞdu since yðtÞ¼0 for t 0
0
es ð1 es Þ
s
¼es LfyðtÞg:
Therefore, we have
Lfyðt þ 1Þg ¼ es LfyðtÞg:
Therefore, we get
e
su
1
Lfyðt þ 2Þg Lfyðt þ 1Þg LfyðtÞg ¼ 0:
Lfyðt þ 2Þg ¼ e2s LfyðtÞg Z0
LfyðtÞgþes LfyðtÞg ¼
s
e ð1 e Þ
s
s
1
s1
or
1
ðs 1Þ ð1 þ es Þ
1
½1 es þ e2s e3s þ . . .
¼
ðs 1Þ
1
X
ð1Þn ens
:
¼
s1
n¼0
¼
LfyðtÞg
or
es ð1es Þ
sðe2s es 1Þ
2
3
1
1
pffiffi
es ð1es Þ 4 pffiffi5
5
pffiffi pffiffi 5
¼
1þ 5
1 5
s
s
s
e2
e2
2 0
13 Hence
1es 4 1 @
1
1
A5
pffiffi
pffiffi
pffiffiffi
¼
yðtÞ
s
5 1 1þ 5 es 1 1 5 es
2
2
" ( pffiffiffi½t )
( pffiffiffi½t )#
1
1þ 5
1 5
¼ pffiffiffi L
L
: EXAMPLE 7.37
2
2
5
LfyðtÞg ¼
¼
½t
X
ð1Þn etn :
n¼0
Solve the differential-difference equation
Hence
"
pffiffiffin pffiffiffin #
1
1þ 5
1 5
an ¼ pffiffiffi
;n
2
2
5
y0 ðtÞ yðt 1Þ ¼ t;
0:
yðtÞ þ yðt1Þ ¼ e;
¼ Lftg:
Now
yðtÞ ¼ 0;
t 0:
Lfy0 ðtÞg
¼ s LfyðtÞgyð0Þ ¼ s LfyðtÞg
and
Lfyðt 1Þg
Solution. Taking Laplace transform of both sides of
the given equation, we get
LfyðtÞg þ Lfyðt1Þg
t 0:
Solution. Taking Laplace transform of both sides,
we have
Lfy0 ðtÞgLfyðt1Þg
EXAMPLE 7.36
Solve the difference equation
t
yðtÞ ¼ 0;
¼ Lfe g:
t
¼ es LfyðtÞg:
Therefore,
fes þ sgLfyðsÞg
¼
1
s2
Applications of Laplace Transform
n
7.21
But
or
1
1
LfyðtÞg ¼ 2 s
¼ s s ðe þ sÞ s3 1 þ es
1
es e2s e3s
þ 2 3 þ ...
¼ 3 1
s
s
s
s
s
2s
3s
1 e
e
e
¼ 3 4 þ 5 6 þ ...
s
s
s
s
1 ns
X
e
¼
:
nþ3
s
n¼0
But
(
ns
ðtnÞnþ2
e
1
¼ ðnþ2Þ ! for t n
L
snþ3
0
otherwise :
Therefore, if [t] denotes the greatest integer less
than or equal to t, then
½t
X
ðt nÞnþ2
yðtÞ ¼
:
ðn þ 2Þ !
n¼0
L 1
en s
snþ4
¼
ðt nÞ2nþ3
:
ð2n þ 3Þ !
Hence
yðtÞ
¼2
½t
X
ðt nÞ2nþ3
n¼0
7.4
ð2n þ 3Þ !
:
INTEGRAL EQUATIONS
Equations of the form
Zb
f ðtÞ ¼ gðtÞ þ
Kðt; uÞ f ðuÞ du
a
and
Zb
gðtÞ ¼
Kðt; uÞ f ðuÞ du;
a
where the function f (t) to be determined appears
under the integral sign are called integral equations.
EXAMPLE 7.38
In an integral equation, K(t, u) is called the
Solve the differential-difference equation
kernel. If a and b are constants, the equation is
y00 ðtÞyðt1Þ ¼ f ðtÞ; yðtÞ ¼ 0; y0 ðtÞ ¼ 0 for t 0; called a Fredholm integral equation. If a is a constant and b = t, then the equation is called a Volterra
0 for t 0
integral equation.
f ðtÞ ¼
2t for t > 0:
If the kernel K(t, u) is of the form K(t u), then
Rt
the integral Kðt uÞ f ðuÞ du represents convoluSolution. Taking Laplace transform of both sides, we
0
tion. Thus, we have
get
Zt
Lfy00 ðtÞg Lfyðt 1Þg ¼ Lf f ðtÞg
f ðtÞ ¼ gðtÞ þ Kðt uÞf ðuÞdu ¼ gðtÞ þ KðtÞ f ðtÞ:
or
2
2
s
0
s LfyðtÞgsyð0Þyð0Þ e LfyðtÞg ¼ 2
s
Such equations are called convolution-type integral
or
2
equations. Taking Laplace transform of convolu2
s
ðs e ÞLfyðtÞg ¼ 2
s
tion-type integral equation, we have
or
Lf f ðtÞg ¼ LfgðtÞg þ LfKðtÞ f ðtÞg
2
2
¼ LfgðtÞg þ LfKðtÞg Lf f ðtÞg;
¼ LfyðtÞg ¼ 2 2
s s ðs es Þ s4 1 es2
by using Convolution theorem. Hence
2
es e2s e3s
ð1 LfKðtÞg Lf f ðtÞgÞ ¼ LfgðtÞg;
¼ 4 1 þ 2 þ 4 þ 6 þ ...
s
s
s
s
which implies
1 es e2s e3s
¼ 2 4 þ 6 þ 8 þ 10 þ . . .
LfgðtÞg
s
s
s
s
:
Lf f ðtÞg ¼
1 LfKðtÞg
1
n s
X
e
:
¼2
Taking inverse Laplace transform yields the solu2nþ4
s
n¼0
tion f (t).
7.22
n
Engineering Mathematics-II
Solution. The given equation in convolution form is
EXAMPLE 7.39
Solve the integral equation
Zt
t
f ðtÞ ¼ e þ sinðt uÞf ðuÞ du:
f ðtÞ f ðtÞ ¼ 16 sin 4 t:
Taking Laplace transform, we get
Lf f ðtÞ f ðtÞg ¼ 16 Lfsin 4tg
0
Solution. Taking Laplace transform of both sides of
the given equation, we get
Lf f ðtÞg ¼ Lfet g þ Lfsin tg Lf f ðtÞg;
or
Lf f ðtÞgLf f ðtÞg ¼ 16 Lfsin 4tg
ðusing convolution theoremÞ:
or
which yields
Lfet g
s2 þ 1
:
Lf f ðtÞg ¼
¼ 2
1 Lfsin tg s ðs þ 1Þ
Using partial fractions, we obtain
2
1 1
Lf f ðtÞg ¼
þ 2 :
sþ1 s
s
Taking inverse Laplace transform yields
½Lf f ðtÞg2 ¼
16ð4Þ
64
¼ 2
:
2
s þ 16 s þ 16
or
8
Lf f ðtÞg ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi :
2
s þ 16
Taking inverse Laplace transform yields
8 J0 ð4tÞ;
f ðtÞ ¼
f ðtÞ ¼ 2et þ t 1:
where J0 is Bessel?s function of order zero.
EXAMPLE 7.40
Solve the integral equation
Zt
f ðtÞ ¼ 1 þ sinðt uÞ f ðuÞ du:
EXAMPLE 7.42
Solve the integral equation
Zt
gðxÞ ¼ f ðxÞ etu f ðuÞ du:
0
Solution. We have
f ðtÞ ¼ 1 þ sin t
0
f ðtÞ:
Solution. The given equation, in convolution form, is
gðtÞ ¼ f ðtÞ et
Taking Laplace transform yields
Lf f ðtÞg ¼ Lf1g þ Lf f ðtÞgLfsin tg
Taking Laplace transform of both sides, we get
or
Lf1g
1
Lf f ðtÞg ¼
¼ 1 Lfsin tg s 1 1
s2 þ1
s2 þ 1 1 1
¼ þ 3:
s3
s s
Taking inverse Laplace transform, we get
t2
f ðtÞ ¼ 1 þ :
2
¼
LfgðtÞg ¼ Lf f ðtÞg Lfet g Lf f ðtÞg
or
LfgðtÞg
ðs 1Þ LfgðtÞg
¼
1 Lfet g
ðs 2Þ
LfgðtÞg
¼ LfgðtÞg þ
s2
¼ LfgðtÞg þ LfgðtÞ Lfe2t gg:
Lf f ðtÞg ¼
Taking inverse Laplace transform yields
EXAMPLE 7.41
Solve
Zt
f ðuÞ f ðt uÞ du ¼ 16 sin 4t:
0
f ðtÞ:
f ðtÞ ¼ gðtÞ þ gðtÞ e2t
Zt
¼ gðtÞ þ gðuÞ e2ðtuÞ du:
0
Applications of Laplace Transform
Definition 7.1. The convolution-type integral equation of the form
Zt
f ðuÞ
du ¼ gðtÞ; 0 < n < 1
ðt uÞn
0
is called Abel?s integral equation.
We consider below examples of this type of integral equations.
EXAMPLE 7.43
Solve the integral equation
Zt
1
2
1 þ 2t t ¼ f ðuÞ pffiffiffiffiffiffiffiffiffiffi du:
tu
n
7.23
Solution. Proceeding as in Example 7.43 above, we
have
1
1
1
2
Lf f ðtÞg ¼ pffiffiffi 1=2 þ 3=2 þ 5=2 ;
s
s
s
which on inversion yields
1
8
f ðtÞ ¼ ½t1=2 þ 2t1=2 þ t3=2 :
3
EXAMPLE 7.45 (Tautochrone Curve)
A particle (bead) of mass m is to slide down a
frictionless curve such that the duration T0 of descent due to gravity is independent of the starting
point. Find the shape of such curve (known as
Tautochrone curve).
0
Solution. The given equation is a special case of
Abel?s integral equation. The convolution form of
this equation is
1
1 þ 2t t2 ¼ f ðtÞ pffi :
t
Solution. Velocity of the bead at the starting point is
zero since it starts from rest at that point, say P with
co-ordinates (x, y). Let Q = (x, u) be some intermediate point during the motion. Let the origin O be
the lowest point of the curve (Figure 7.12). Let the
length of the arc OQ be s.
Taking Laplace transform yields
1
Lf f ðtÞg L pffi
t
or
y
¼ Lf1g þ 2Lftg Lft g
2
P(x, y)
rffiffiffi
1 2
2
¼ þ 2 3
Lf f ðtÞg
s s s
s
y
or
Q(x, u )
u
1
1
2
2
Lf f ðtÞg ¼ pffiffiffi 1=2 þ 3=2 5=2 :
s
s
s
0
Figure 7.12
Taking inverse transform, we get
1
t1=2
2t1=2
2t3=2
þ
f ðtÞ ¼ pffiffiffi
ð1=2Þ ð3=2Þ ð5=2Þ
1 t1=2
2t1=2
2t3=2
pffiffiffi pffiffiffi
¼ pffiffiffi pffiffiffi þ
ð1=2Þ ð3=2Þð1=2Þ 1
¼ ½t1=2 þ 4t1=2 8t3=2 :
EXAMPLE 7.44
Solve the integral equation
Zt
1
f ðuÞ pffiffiffiffiffiffiffiffiffiffi ¼ 1 þ t þ t2 :
tu
0
x
By law of conservation of energy, potential energy
at P + kinetic energy at P = potential energy at
Q + Kinetic energy at Q, that is,
2
1
ds
;
mgy þ 0 ¼ mgu þ m
2
dt
where ds
dt is the instantaneous velocity of the particle
at Q.
Thus
2
ds
¼ 2gðy uÞ
dt
and so
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ds
¼ 2gðy uÞ;
dt
7.24
n
Engineering Mathematics-II
negative sign since s decreases with time. The total
time T0 taken by the particle to go from P to Q is
ZT0
T0 ¼
Z0
dt ¼
y
0
If
ds
du
ds
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ¼
2gðy uÞ
Zy
0
ds
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi:
2gðy uÞ
= f (u), then ds = f (u) du and so
Zy
1
f ðuÞ
T0 ¼ pffiffiffiffiffi
pffiffiffiffiffiffiffiffiffiffiffi du
yu
2g
0
The convolution form of this integral equation is
1
1
T0 ¼ pffiffiffiffiffi f ðyÞ pffiffiffi :
y
2g
Taking Laplace transform of both sides and using
Convolution theorem, we have
1
1
LfT0 g ¼ pffiffiffiffiffi Lf f ðyÞg L pffiffiffi
y
2g
or
pffiffiffiffiffiffiffiffiffiffiffi
pffiffiffiffiffi
2g=
2g T0 =s
C0
Lf f ðyÞg ¼ pffiffiffiffiffiffiffiffi ¼ 1=2 T0 ¼ 1=2 ;
s
s
T =s
where C0 is a constant. Inverse Laplace transform
then yields
C
f ðyÞ ¼ pffiffiffi :
y
Since f(y) =
ds
dy
¼
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
ffi
2
1 þ dx
dy , we get
2
dx
C2
1þ
¼
y
dy
or
2
dx
C2
C2 y
1¼
¼
y
dy
y
or
sffiffiffiffiffiffiffiffiffiffiffiffiffiffi
dx
C2 y
¼
dy
y
or
ffi
Z sffiffiffiffiffiffiffiffiffiffiffiffiffi
C2 y
x¼
dy:
y
Putting y = C2 sin2 h2, we get
C2
C2
x¼
ðh þ sin hÞ; y ¼
ð1 cos hÞ;
2
2
which are the parametric equations of a cycloid.
7.5
INTEGRO-DIFFERENTIAL EQUATIONS
An integral equation in which various derivatives of
the unknown function f(t) are also present is called
an integro-differential equation. These types of
equations can also be solved by the method of
Laplace transform.
EXAMPLE 7.46
Solve the following integro-differential equation:
Zt
0
y ðtÞ ¼ yðuÞ cosðt uÞ du; yð0Þ ¼ 1:
0
Solution. We write the given equation in convolution
form as
y0 ðtÞ ¼ yðtÞ cos t:
Taking Laplace transform and using Convolution
theorem yields
Lfy0 ðtÞg ¼ LfyðtÞg Lfcos tg
or
sLfyðtÞg yð0Þ ¼ LfyðtÞg
or
s
s 2
LfyðtÞg ¼ 1;
s þ1
or
LfyðtÞg ¼
s2
s
þ1
since yð0Þ ¼ 1
s2 þ 1 1 1
¼ þ 3:
s3
s s
Taking inverse Laplace transform, we get
1
yðtÞ ¼ 1 þ t2 :
2
EXAMPLE 7.47
Solve
Zt
y0 ðtÞ þ 5 yðuÞ cos 2ðt uÞ dy ¼ 10; yð0Þ ¼ 2:
0
Solution. Convolution form of the equation is
y0 ðtÞ þ 5 cos t yðtÞ ¼ 10:
Taking Laplace transform and using Convolution
theorem, we have
5sLfyðtÞg 10
¼
sLfyðtÞg yð0Þ þ 2
s þ4
s
Applications of Laplace Transform
7.25
(b) Integrating by parts, we get
or
2s3 þ 10s2 þ 8s þ 40
s2 ðs2 þ 9Þ
1 8 40
10s
50
þ
¼
þ þ
:
9 s s2 s2 þ 9 s2 þ 9
LfyðtÞg ¼
Hence
yðtÞ ¼
7.6
n
1
50
8 þ 40t þ 10 cos t þ sin 3t :
9
3
SOLUTION OF PARTIAL DIFFERENTIAL
EQUATION
@u
L
@t
Z1
¼
est
0
ZT
@uðx; tÞ
dt
@t
@uðx; tÞ
dt
@t
9
80 "
#T
ZT
=
<
¼ lim
est uðx; tÞ þ s est uðx; tÞdt
T!0:
;
¼ lim
T!0
est
0
ZT
0
¼s
est uðx; tÞdt uðx; 0Þ
Consider the function u = u(x, t), where t 0 is a
0
time variable. Suppose that u(x, y), when regarded
¼ sUðx; sÞ uðx; 0Þ:
as a function of t, satisfies the sufficient conditions
@u
for the existence of its Laplace transform.
ðcÞ Taking V ¼
; we have byðaÞ;
@x
Denoting the Laplace transform of u(x, t) with
respect to t by U(x, s), we see that
@2u
@V
d
Z1
¼L
L
¼ ðVðx; sÞÞ
2
@x
@x
dx
est uðx; tÞ dt:
Uðx; sÞ ¼ Lfuðx; tÞg ¼
d d
d2
0
ðU ðx; sÞÞ ¼ 2 ðUðx; sÞÞ:
¼
dx
dx dx
The variable x is the untransformed variable. For
@u
(d) Let v = @t . Then
example,
@2u
@v
aðxtÞ
ax
at
ax 1
¼L
¼ sVðx; sÞ vðx; 0Þ
L
g ¼ e L fe g ¼ e
:
Lfe
2
@t
@t
sþa
@u
¼ s½sUðx; sÞ uðx; 0Þ ðx; 0Þ
Theorem 7.2. Let u(x, t) be defined for t 0. Then
@t
@u
d
@u
(a) L
¼ ðUðx; sÞÞ
¼ s2 Uðx; sÞ uðx; 0Þ ðx; 0Þ:
@x
dx
@t
Theorem
7.2
suggest
that
if
we
apply
Laplace
trans@u
¼ s Uðx; sÞ uðx; 0Þ
(b) L
form
to
both
sides
of
the
given
partial
differential
@t
equation, we shall get an ordinary differential equa@2u
d2
tion in U as a function of single variable x. This
(c) L
¼ 2 ðUðx; sÞÞ
dx
@x2
ordinary differential equation is then solved by the
2
usual methods.
@ u
@u
¼ s2 Uðx; sÞ s uðx; 0Þ ðx; 0Þ:
(d) L
2
@t
@t
EXAMPLE 7.48
Proof: (a) We have, by Leibnitz?s rule for differ- Solve
entiating under the integration,
@u @u
Z1
Z1
¼
; uðx; 0Þ ¼ x; uð0; tÞ ¼ t;
@u
@u
d
@x @t
est
est uðx; tÞ dt
¼
dt ¼
L
@x
@x
dx
Solution. Taking Laplace transform, we get
0
0
d
@u
@u
¼ ðUðx; sÞÞ:
¼L
:
L
dx
@x
@t
7.26
n
Engineering Mathematics-II
Using Theorem 7.2, we get
d
½U ðx; sÞ ¼ s U ðx; sÞ uðx; 0Þ ¼ s U ðx; sÞ x:
dx
Solution. Taking Laplace transform with respect to t,
we get
@u
@u
þL x
@t
@x
L
Thus, we have first order differential equation
d
½Uðx; sÞ s Uðx; sÞ ¼ x
dx
The integrating factor is
R
e s dx ¼ esx :
Therefore,
U ðx; sÞ esx ¼
Z
sUðx; sÞ uðx; 0Þ þ x
d
x
Uðx; sÞ ¼ :
dx
s
d
s
1
Uðx; sÞ þ Uðx; sÞ ¼ :
dx
x
s
xesx dx
esx
esx
dx þ C
s
s
x esx esx
þ 2 þC
¼
s
s
(constant of integration):
¼ x
This yields
x 1
Uðx; sÞ ¼ þ 2 þ C esx : ð12Þ
s s
Now the boundary condition u(0, t) is a function of t.
Taking Laplace transform of this function, we have
1
Uð0; sÞ ¼ Lfuð0; tÞg ¼ Lftg ¼ 2 :
s
ð13Þ
The integrating factor is
Rs
dx
e x
¼ es log x ¼ xs :
Therefore solution of (13) is
Z
1
1 xsþ1
xsþ1
Uðx; sÞxs ¼
þC ¼
þC
xs dxþC ¼
sðsþ1Þ
s
s sþ1
and so
U(x, s) = x + C (constant of integration). (14)
sðsþ1Þ
Now since U(0, t) = 0, its Laplace transform is
0, that is, U(0, s) = 0. Therefore, (14) implies C = 0.
Hence
x
1
1
¼x :
Uðx; sÞ ¼
sðs þ 1Þ
s sþ1
Taking inverse Laplace transform, we get the solution as
uðx; tÞ ¼ xð1 et Þ:
Then taking x = 0 in (12), we have
1
1
¼ 2þC
2
s
s
and so C = 0. Thus, we have
x 1
Uðx; sÞ ¼ þ 2 :
s s
Taking inverse Laplace transform, we have
uðx; tÞ ¼ x þ t:
EXAMPLE 7.50
Solve
@u @ 2 u
;
¼
@t @x2
under the conditions
x > 0;
t>0
uðx; 0Þ ¼ 1; uð0; tÞ ¼ 0 and lim uðx; tÞ ¼ 1:
EXAMPLE 7.49
Solve the partial differential equation
x > 0;
which yields
Since u(x, 0) = 0, this reduces to
Z
@u
@u
þx
¼ x;
@t
@x
¼ Lfxg
t>0
with the initial and boundary conditions u(x, 0) = 0,
x > 0 and u(0, t) = 0 for t > 0.
x!1
Solution. The given equation is heat conduction
equation in a solid, where u(x, t) is the temperature at
position x at any time t and diffusivity is 1. The
boundary condition u(0, t) = 0 indicates that temperature at x = 0 is 0 and lim uðx; tÞ ¼ 1 indicates
x!1
that the temperature for large values of x is 1 whereas
Applications of Laplace Transform
u(x, 0) = 1 represents the initial temperature 1 in the
semi-infinite medium (x > 0) (Figure 7.13).
n
7.27
subject to the conditions
uðx; 0Þ ¼ 0; x > 0;
uð0; tÞ ¼ t; t > 0 and lim uðx; tÞ ¼ 0:
x!1
Solution. Taking Laplace transform, we have
0
The solution of this equation is
pffi
pffi
U ðx; sÞ ¼ c1 e s x þ c2 e s x
Figure 7.13
Taking Laplace transform, yields
d2
sUðx; sÞ uðx; 0Þ ¼ 2 Uðx; sÞ:
dx
Since u(x, 0) = 1, we have
d2
Uðx; sÞ s Uðx; sÞ ¼ 1:
dx2
The general solution of this equation is
Uðx; sÞ ¼ C:F: þ P:I:
pffi
pffi
1
¼ ½c1 e s x þ c2 e sx þ :
s
The conditions u(0, t) = 0 yields
U ð0; sÞ ¼ Lfuð0; tÞg ¼ 0;
x!1
lim Uðx; sÞ ¼ lim Lfuðx; tÞg
x!1
x!1
Therefore, c1 = 0 and (18) reduces to
pffi
Uðx; sÞ ¼ c2 e s x :
ð19Þ
Also, since u(0, t) = t, we have
ð15Þ
¼
Uð0; sÞ
ð16Þ
lim Uðx; sÞ ¼ lim Lfuðx; tÞg ¼ Lf lim uðx; tÞg
x!1
1
ð17Þ
¼ Lf1g ¼ :
s
Now (15) and (17) imply c1 = 0. Then (16) implies
c2 = 1s . Hence
pffi
1 e sx
:
Uðx; sÞ ¼ s
s
Taking inverse Laplace
we get
( ptransform,
ffi )
sx
e
uðx; tÞ ¼ 1 L1
s
x
x
¼ erf pffi :
¼ 1 1 erf pffi
2 t
2 t
EXAMPLE 7.51
Solve
x!1
¼ Lf lim uðx; tÞg ¼ Lf0g ¼ 0ðfiniteÞ:
x!1
x!1
ð18Þ
Since lim u(x, t) = 0, we have
whereas lim u(x, t) = 1 yields
x!1
d2
Uðx; sÞ s Uðx; sÞ ¼ 0:
dx2
x
x
1
:
s2
Hence, (19) yields c2 = s12 . Thus
Uðx; sÞ
pffi
Since L1 {e s x } =
theorem, we have
uðx; tÞ
¼
1 pffis x
e
:
s2
x2
pxffiffiffiffiffi e 4t , by Convolution
2 t3
¼
Zt
x
x2
ðt uÞ pffiffiffiffiffiffiffiffi e 4u du:
2 u3
0
Putting l =
2
x
4u,
uðx; tÞ
we get
2
¼ pffiffiffi
Z1
x2
l2
e
t 2 dl:
4l
pffi
x=2 t
EXAMPLE 7.52
Solve
@u
@2u
¼2 2
@t
@x
@u @ 2 u
¼
; x > 0; t > 0
@t @x2
subject to the conditions u(0, t) = 0, u(5, t) = 0,
u(x, 0) = sin x.
7.28
n
Engineering Mathematics-II
Solution. Taking Laplace transform and using u(x, 0)
= sin x, we get
d2
s
1
Uðx; sÞ Uðx; sÞ ¼ sin x:
dx2
2
2
Complementary function for this equation is
ffi
pffiffiffi
pffis
s
c1 e 2 x þ c2 e 2 x and particular integral is
1
2ð2 þðs=2ÞÞ sin x. Thus the complete solution is
pffis
pffis
1
sinx:
Uðx;sÞ ¼ c1 e 2 x þ c2 e 2 x þ
2ð2 þ ðs=2ÞÞ
ð20Þ
Since u(0, t) = 0, we have U(0, t) = 0 and since
u(5, t) = 0, U(5, t) = 0. Therefore,
(20) gives
pffiffiffiffiffi
pffi
5 s=2
c1 þ c2 ¼ 0
and c1 e
þ c2 e5ð s=2Þ ¼ 0:
These relations imply c1 = c2 = 0. Hence
1
1
sin x ¼
sin x:
Uðx; sÞ ¼
2
2ð þ s=2Þ
s þ 22
Taking inverse Laplace transform, we get
2
Uðx; tÞ ¼ e2 t sin s:
EXAMPLE 7.53
Solve one-dimensional wave equation
2
@2u
2@ y
¼
a
;
x > 0;
t>0
@t2
@x2
subject to the condition y(x, 0) = 0, x > 0; yt(x, 0) = 0,
x > 0, y(0, t) = sin vt and lim y(x, t) = 0.
x!1
Solution. The displacement is only in the vertical
direction and is given by y(x, t) at position x and time
qffiffiffi
t. For a vibrating string, the constant a equals T ,
where T is tension in the string and is mass per
unit length of the vibrating string (Figure 7.14).
Taking Laplace transform, we get
d2
s2 Yðx; sÞ syðx; 0Þ yt ðx; 0Þ a2 dx
2 Yðx; sÞ ¼ 0
or
d2
s2
Yðx;
sÞ
ds2
a2
¼0
ð21Þ
The general solution of (21) is
Yðx; sÞ ¼ c1 ea x þ c2 e a x
s
s
ð22Þ
The condition lim y(x, t) implies c1 = 0.
x!1
Since y(0, t) = sin vt, we have
v
:
Yð0; sÞ ¼ fyð0;tÞg ¼ 2
s þ v2
v
Therefore, (22) implies c2 ¼ s2 þv
2 and so
¼
Yðx; sÞ
v
s
e a x :
s2 þ v2
Taking inverse Laplace transform, we have
(
sin v t ax
for t > ax
yðx; tÞ ¼
0
for t < ax :
x
x
¼ sin v t H t :
a
a
EXAMPLE 7.54
Solve
@2y @2y
¼
;
@t2 @x2
for 0 <x< 1;
t> 0
subject to y(x, 0) = 0, 0 < x < 1 ; y(0, t) = 0, t > 0,
y(1, t) = 0, t > 0 and yt(x, 0) = x, 0 < x < 1.
Solution. Taking Laplace transform and using y(x, 0)
= 0 and yt(x, 0) = x, we get
d 2 Yðx; sÞ
s2 Yðx; sÞ ¼ x;
dt2
whose solution is given by
y
Yðx; sÞ ¼ c1 cosh sx þ c2 sinh sx y(x , t)
x
0
Figure 7.14
x
:
s2
Now y(0, t) = 0 implies that Y(0, s) = 0 and so c1 = 0.
Similarly, y(1, t) = 0 implies Y(1, s) = 0 and so
1
c2 sinh s s12 ¼ 0: Thus c2 ¼ s2 sinh
x : Hence
Yðx; sÞ ¼
s2
1
x
: sinh sx 2 :
sinh x
s
Applications of Laplace Transform
This function has simple poles at ni, n = ±1,
±2, . . ., and a pole of order 2 at s = 0. Now
sinh sx
ResðniÞ ¼ lim ðs niÞets : 2
s!n i
s sinh s
ðs n iÞ
sinh sx
¼ lim
lim ets
s!n i sinh s s!n i
s2
n i t
1
e
sinh nix
¼
:
cosh n i
n2 2
nþ1
ð1Þ
¼
en i t sin nx;
n2 2
Resð0Þ ¼ xt:
Hence, by Complex inversion formula,
1
X
ð1Þnþ1 n i t
e
sin nx xt
yðx; tÞ ¼ xt þ
n2 2
n¼1
¼
7.7
1
2X
ð1Þnþ1
sin nx sin nt:
2 n¼1 n2
Setting s = 1, we get
Z1
sin t
et
dt ¼ tan1 1 ¼ :
t
4
0
Further, letting s ! 0 in (23), we get
Z1
sin t
dt ¼ tan1 1 ¼ :
t
2
0
EXAMPLE 7.56
Evaluate the integral
Z1
sin tx
dx:
xð1 þ x2 Þ
0
Solution. Let
R1 sin tx
f ðtÞ ¼ xð1þx
2 Þ dx:
0
0
Laplace transforms can be used to evaluate certain
integrals. In some cases the given integral is a special
case of a Laplace transform for a particular value of
the transform variable s. To evaluate an integral
containing a free parameter, we first take Laplace
transform of the integrand with respect to the free
parameter. The resulting integral is then easily
evaluated. Then we apply inverse Laplace transform to get the value of the given integral. In some
cases, Theorem 5.9, regarding Laplace transform is
used to evaluate the given integral.
t
0
¼
dx
xð1 þ
0
x2 Þ ðs2
þ x2 Þ
1
1
dx
1 þ x2 s2 þ x2
0
1 1
s1
¼ 2
¼ 2
s 1 2 2s
2s 1
s
1
1
1
¼
:
¼
2 sðs þ 1Þ
2 s sþ1
1
¼ 2
s 1
Z1 f ðtÞ ¼ 2 ð1 et Þ:
EXAMPLE 7.57
Evaluate
0
dt ¼
0
Z1
Taking inverse Laplace transforms, we get
EXAMPLE 7.55
Evaluate the integral
Z1
sin t
et
dt;
I¼
t
and show that
7.29
Taking Laplace transform with respect to t, we have
Z1
Z1
1
dx est sin tx dx
FðsÞ ¼
xð1 þ x2 Þ
EVALUATION OF INTEGRALS
R1 sin t
n
Z1
2:
sin2 tx
dx:
x2
0
Solution. We know that
Z1
sin t
sin
1
¼
dt ¼ tan1 :
est
L
t
t
s
0
ð23Þ
Solution. We have
Z1 2
Z1
sin tx
1 cosð2txÞ
f ðtÞ ¼
dx ¼
dx:
x2
2x2
0
0
7.30
n
Engineering Mathematics-II
Taking Laplace transform with respect to t, we have
FðsÞ ¼
1
2
Z1
0
¼
1
s
Z1
Z1
1 1
s
2
dx
dx
¼
x2 s 4x2 þ s2
s 4x2 þ s2
dy
1 h 1 yi1
¼
tan
¼
y2 þ s 2 s2
s 0
0
:
2s2
0
Thus, taking inverse Laplace transformation, we get
t t
f ðtÞ ¼
¼ sgn t:
2
2
EXAMPLE 7.58
Evaluate
Taking 2m 1 = 12 and 2n 1 = 12 , we get m = 34
and n = 14. Hence using the relation G( p) G(1 p) =
sin p, 0 < p < 1, we have
pffiffiffi
1 ð3=4Þð1=4Þ
1 pffiffiffi 2
¼ pffiffi 2 ¼ pffiffi :
Lf f ðtÞg ¼ pffiffi
4 s
ð1Þ
4 s
4 s
Taking inverse Laplace transform yields
pffiffiffi pffiffiffiffiffiffi
2 1=2
2 t1=2
pffiffiffi ¼
t
:
f ðtÞ ¼
4
4
Putting t = 1, we get
rffiffiffi
Z1
1 sin x2 dx ¼
2 2
0
and so
Z1
Z1
sin x2 dx:
1
1
Solution. Let
R1
f ðtÞ ¼ sin t x2 dx
EXAMPLE 7.59
Evaluate the integral
0
Taking Laplace transform, we get
Z1
Z1
st
Lf f ðtÞg ¼
e dt sin t x2 dx
0
Z1
Z1
¼
dx
0
Z1
0
est sin tx2 dt
0
2
Z=2
0
1
¼ pffiffi
2 s
s3=2 tanhðtanhÞ1=2 sec2 hdh
s2 ð1þtan2 hÞ
Z=2 pffiffiffiffiffiffiffiffiffiffi
Z=2
1
tanh dh¼ pffiffi sin1=2 hcos1=2 dh:
2 s
0
0
But we know that
Z=2
1
ðmÞðnÞ
sin2m1 hcos2n1 hdh ¼ bðm;nÞ ¼
:
2
2ðmþnÞ
0
cos tx
dx;
x2 þ 1
Solution. Let
Z1
f ðtÞ ¼
Z1
x2
dx:
s2 þ x4
0
0
pffiffipffiffiffiffiffiffiffiffiffiffiffiffi
Put x2 = s tan h, that is, x = s tan h: Then
pffiffi 1
dx ¼ s: ðtan hÞ1=2 sec2 h dh:
2
Therefore,
1
Lff ðtÞg¼
2
Z1
t> 0:
0
Lfsin tx g dx ¼
¼
rffiffiffi
:
sin x dx ¼
2
2
cos tx
dx:
x2 þ 1
0
Taking Laplace
with
8 transform
9 respect to t, we get
<Z1 cos tx = Z1
cos tx
Lf f ðtÞg ¼ L
dx ¼
dx
L 2
: x2 þ 1 ;
x þ1
Z1
¼
0
ðx2
0
Z1
¼s
0
s
dx
þ 1Þ ðs2 þ x2 Þ
dx
ðx2 þ 1Þ ðs2 þ x2 Þ
0
1
1
dx
x2 þ 1 s2 þ x2
0
s h 1
xi1
¼ 2
tan x tan1
s 1
s 0
s =2
¼ 2
:
¼
s 1 2 2s
sþ1
s
¼ 2
s 1
Z1 Applications of Laplace Transform
Now, taking inverse Laplace transform, we have
f ðtÞ ¼ et ; t> 0:
2
EXAMPLE 7.60
Evaluate
EXAMPLE 7.62
Evaluate
Z1
pffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
erf u erf ðt uÞ du:
Solution. Let
where J0 is Bessel?s function of order zero.
Solution. We know that
1
LfJ0 ðtÞg ¼ pffiffiffiffiffiffiffiffiffiffiffiffi :
2
s þ1
Therefore,
d
d
1
pffiffiffiffiffiffiffiffiffiffiffiffi
fLfJ0 ðtÞgg ¼ ds
ds
s2 þ 1
2s
¼
:
ðs2 þ 1Þ3=2
LftJ0 ðtÞg ¼ But, by definition
Z1
LftJ0 ðtÞg ¼
est t J0 ðtÞ dt:
est t J0 ðtÞ dt ¼ 0
2s
ðs2
þ 1Þ3=2
:
Taking s = 0, we get
Z1
t J0 ðtÞ dt ¼ 0:
pffiffiffiffiffiffiffiffiffiffi
pffiffiffi
u erf t u du:
Then, by Convolution theorem, we have
pffi
pffi
FðtÞ ¼ Lferf tgLferf tg
1
1
1
¼ pffiffiffiffiffiffiffiffiffiffiffi : pffiffiffiffiffiffiffiffiffiffiffi ¼ 2
ðs
þ 1Þ
s
s sþ1 s sþ1
1 1
1
¼
þ 2þ
:
s
s
sþ1
Taking inverse transform, we get
f ðtÞ ¼ 1þtþet :
MISCELLANEOUS EXAMPLES
EXAMPLE 7.63
Solve the differential equation, using Laplace
transform y00 þ 4y0 þ 4y ¼ et given that yð0Þ = 0
and y0 ð0Þ ¼ 0.
Solution. Taking Laplace transform, we have
Lfy00 ðtÞg þ 4Lfy0 ðtÞg þ 4LfyðtÞg ¼ Lfet g
or
s2 Y ðsÞ syð0Þ y0 ð0Þ þ 4fsY ðsÞ yð0Þg þ 4Y ðsÞ
0
EXAMPLE 7.61
Evaluate
erf
0
7.8
0
Zt
f ðtÞ ¼
0
Z1
7.31
0
t J0 ðtÞ dt;
Hence
Zt
n
1
:
sþ1
Using the initial conditions, we have
¼
Z1
pffi
e2t erf tdt:
s2 Y ðsÞ þ 4sY ðsÞ þ 4Y ðsÞ ¼
0
Solution. We have
pffi
Lferf tg ¼
or
Z1
0
pffi
1
est erf tdt ¼ pffiffiffiffiffiffiffiffiffiffiffi :
s sþ1
Taking s = 2, we get
Z1
pffi
1
e2t erf tdt ¼ pffiffiffi :
2 3
0
1
sþ1
1
1
¼
þ 4s þ 4Þ ðs þ 1Þðs þ 2Þ2
ðs þ
1
1
1
¼
:
s þ 1 s þ 2 ðs þ 2Þ2
Y ðsÞ ¼
1Þðs2
Taking inverse Laplace transform, we get
yðtÞ ¼ et e2t te2t :
7.32
n
Engineering Mathematics-II
EXAMPLE 7.64
Using Laplace transform, solve
ðD2 þ 5D 6Þy ¼ x2 ex ; yð0Þ ¼ a; y0 ð0Þ ¼ b:
Using initial conditions xð0Þ ¼ x0 ð0Þ ¼ 0, we get
s2 4
X ðsÞðs2 þ 1Þ ¼
ðs2 þ 4Þ2
Solution. Applying Laplace transform to the given
equation, we get
s2 Lfyg syð0Þ yð0Þ þ 5½sLfyg yð0Þ 6Lfyg
or
¼
ð1Þ4 2 !
ðs þ 1Þ
3
¼
ðs2 1Þðs2 þ 4Þ2
5
3
5
þ
:
¼
9ðs2 þ 4Þ 3ðs2 þ 4Þ2 9ðs2 þ 1Þ
2
ðs þ 1Þ3
or
s2 Lfyg as b þ 5½sLfyg a 6Lfyg ¼
2
ðs þ 1Þ3
or
Lfyg½s2 þ 5s 6 ¼
2
ðs þ 1Þ
þ as þ b þ 5a
3
or
as þ 5a þ b
2
þ
s2 þ 5s 6 ðS þ 1Þ3 ðs2 þ 5s 6Þ
1
1
3
1
19
1
¼
3
2
5 ðs þ 1Þ
50 ðs þ 1Þ
500 ðs þ 1Þ
1
1
2
1
þ
þ
28 ðs 1Þ 875 ðs þ 6Þ
(by partial fractions).
Lfyg ¼
Taking inverse Laplace transform, we have
a b 6x 6a þ b x
1
yðxÞ ¼
e þ
e x2 ex
7
7
50
19 x
1 x
2 6x
e þ e þ
e
:
500
28
875
s2 4
X ðsÞ ¼
Taking inverse Laplace transform, we have
5 sin 2t 8
5
þ ½sin 2t 2t cos 2t sin t:
xðtÞ ¼
9 2
3
9
EXAMPLE 7.66
Solve the differential equation
d2y
dy
2 þ y ¼ e2x ; yð0Þ ¼ 2; y0 ð0Þ ¼ 1
2
dx
dx
by using Laplace transforms.
Solution. Taking Laplace transform, we have
s2 Y ðsÞ syð0Þ y0 ð0Þ 2fsY ðsÞ yð0Þg þ Y ðsÞ
1
:
s2
Since yð0Þ ¼ 2 ; y0 ð0Þ ¼ 1, we have
¼
s2 Y ðsÞ 2s þ 1 2sY ðsÞ þ 4 þ Y ðsÞ ¼
1
s2
or
s2 Y ðsÞ 2sY ðsÞ þ Y ðsÞ ¼ 2s 5 þ
1
s2
or
EXAMPLE 7.65
2
Solve the differential equation dd tx2 þ x ¼ t cos 2t
0
under the conditions: xð0Þ ¼ x ð0Þ ¼ 0.
Solution. The given differential equation is
d2x
þ x ¼ t cos 2t; xð0Þ ¼ x0 ð0Þ ¼ 0:
dt2
Taking Laplace transform of both sides of the given
equation, we have
Lfx00 ðtÞg þ LfxðtÞg ¼ Lft cos 2tg
or
s2 X ðsÞ sxð0Þ x0 ð0Þ þ X ðsÞ ¼
s2 4
ðs2 þ 4Þ2
:
ðs2 2s þ 1ÞY ðsÞ ¼
2s2 7s þ 11
s2
or
Y ðsÞ ¼
2s2 9s þ 11
ðs 2Þðs 1Þ2
1
1
4
þ
¼
:
s 2 s 1 ðs 1Þ2
Taking inverse Laplace transform, we get
(
)
1
2x
x
1
yðxÞ ¼ e þ e 4 L
ðs 1Þ2
¼ e2x þ ex 4ex ðxÞ ¼ ex ð1 4xÞ þ e2x :
Applications of Laplace Transform
EXAMPLE 7.67
Solve the following differential equation using
Laplace transforms:
d3y
d2y
dy
3 2 þ 3 y ¼ t2 et where
3
dt
dt
dt
2 dy
d y
yð0Þ ¼ 1;
¼ 0;
¼ 2:
dt t¼0
dt2 t¼0
EXAMPLE 7.68
Solve the following simultaneous differential
equations:
dy
dx
(a) 3 dx
dt y ¼ 2t ; dt þ dt y ¼ 0 with the condition xð0Þ ¼ yð0Þ ¼ 0:
dy
dx
t
with the
(b) dx
dt þ 4 dt y ¼ 0 ; dt þ 2y ¼ e
condition xð0Þ ¼ yð0Þ ¼ 0:
3x0 y ¼ 2t
x0 þ y 0 y ¼ 0
with conditions
yð0Þ ¼ 1 ; y0 ð0Þ ¼ 0 ; y00 ð0Þ ¼ 2:
Taking Laplace transform, we get
s3 Y ðsÞ s2 yð0Þ sy0 ð0Þ y00 ð0Þ 3½s2 yðsÞ
syð0Þ y0 ð0Þ þ 3½sY ðsÞ yð0Þ Y ðsÞ
2
¼
:
ðs 1Þ3
Using the given initial conditions, we have
s3 Y ðsÞ s2 þ 2 3½s2 Y ðsÞ s þ 3½sY ðSÞ 1
2
Y ðsÞ ¼
ðs 1Þ3
or
with xð0Þ ¼ yð0Þ ¼ 0.Taking Laplace transforms,
we get
2
3fsX ðsÞ xð0Þg Y ðsÞ ¼ 2
s
and
sX ðsÞ xð0Þ þ sY ðsÞ yð0Þ Y ðsÞ ¼ 0:
Using the initial conditions, the above equations
reduce to
2
3sX ðsÞ Y ðsÞ ¼ 2
s
and
sX ðsÞ þ sY ðsÞ yðsÞ ¼ 0
2
ðs 1Þ
3
or
or
3
ðs 1Þ Y ðsÞ ¼
¼
2
ðs 1Þ
2
3
2
ðs 1Þ
þ
6
ðs 1Þ2
ðs 1Þ
3
þ ðs 1Þ s
"
s
ðs 1Þ3
#
1
ðs 1Þ þ 1
¼
þ
ðs 1Þ6 s 1
ðs 1Þ3
2
1
1
1
¼
þ
6
3
s
1
ðs 1Þ
ðs 1Þ
ðs 1Þ3
2
2
s2
ð24Þ
and
Thus
Y ðsÞ ¼
3sX ðsÞ Y ðsÞ ¼
þ s2 3s þ 1
2
ðs 1Þ3
7.33
Solution. (a) The given simultaneous differential
equations are
Solution. The given differential equation is
y000 3y00 þ 3y0 y ¼ t2 et
½s3 3s2 þ 3s 1Y ðsÞ s2 þ 3s 1 ¼
n
Taking inverse Laplace transform, we get
t 5 et
t2
þ et tet et :
y¼
60
2
sX ðsÞ þ ðs 1ÞY ðsÞ ¼ 0
ð25Þ
Multiplying the equation (25) by 3 and then subtracting (24) from it, we get
2
ð3s 2ÞY ðsÞ ¼ 2
s
and so
2
1
3
3
ð26Þ
Y ðsÞ ¼ 2
¼ 2þ s ð3s 2Þ s
2s 2 s 23
(by partial fractions)
Taking inverse Laplace transform, we have
3 3 2t
y ¼ t þ e3 :
2 2
ð27Þ
7.34
n
Engineering Mathematics-II
Substituting the value of Y ðsÞ from (26) in (24), we
get
2
1
3
3
3sX ðsÞ ¼ 2 þ 2 þ s
s
2s 2 s 23
or
5
2
þ 7sðs þ 1Þ 7s s 34
"
#
5 1
1
8
1
1
:
¼
þ
7 s sþ1
21 s 34 s
X ðsÞ ¼
so that
1
1
1
þ 2 3
s
2s
2s s 23
"
#
1
1
3 1
1
¼ 3þ 2
s
2s
4 s 23 s
X ðsÞ ¼
Taking inverse transform, we get
1 5
8 3t
x ¼ et þ e 4 :
3 7
21
Taking inverse Laplace transform, we have
t2 t 3 3 2t
þ þ e3
ð28Þ
2 2 4 4
Thus (27) and (28) provides the solution to the given
system.
(b) Taking Laplace transform, we have
x¼
sX ðsÞ xð0Þ þ 4ðsY ðsÞ yð0Þ Y ðsÞ ¼ 0
and
sX ðsÞ xð0Þ þ 2Y ðsÞ ¼
1
:
sþ1
Using the given initial conditions, we get
sX ðsÞ þ ð4s 1ÞY ðsÞ ¼ 0
ð29Þ
SX ðsÞ þ 2Y ðsÞ ¼ 0
ð30Þ
and
Subtracting (30) from (29), we get
1
ð4s 3ÞY ðsÞ ¼ sþ1
x0 ðtÞ þ 2y0 ðtÞ þ 2xðtÞ þ 2yðtÞ ¼ 0;
where xð0Þ ¼ 1; yð0Þ ¼ 1:
Solution. We want to solve
x0 ðtÞ þ y0 ðtÞ þ xðtÞ ¼ et ;
x0 ðtÞ þ 2y0 ðtÞ þ 2xðtÞ þ 2yðtÞ ¼ 0;
subject to the conditions xð0Þ ¼ 1 ; yð0Þ ¼ 1.
Taking Laplace transform of the given equations,
we have
1
sX ðsÞ xð0Þ þ sY ðsÞ yð0Þ þ X ðsÞ ¼ sþ1
and
sX ðsÞ xð0Þ þ 2½sY ðsÞ yð0Þ þ 2X ðsÞ þ 2Y ðsÞ
¼ 0:
or
1
ðs þ 1Þð4s 3Þ
"
#
1
1
1
:
¼
7 s þ 1 s 34
EXAMPLE 7.69
Solve the simultaneous differential equations using
Laplace transforms:
x0 ðtÞ þ y0 ðtÞ þ xðtÞ ¼ et ;
Y ðsÞ ¼
ð31Þ
Taking inverse Laplace transform, we get
1
3t
y ¼ ½et e 4 :
7
Putting the value of Y ðsÞ from (31) in (30), we get
5
2
sX ðsÞ ¼
þ 7ðs þ 1Þ 7 s 34
Using the given initial conditions, we get
1
sX ðsÞ þ X ðsÞ þ sY ðsÞ ¼ sþ1
and
sX ðsÞ þ 1 þ 2sY ðsÞ 2 þ 2X ðsÞ þ 2Y ðsÞ ¼ 0
that is,
1
sþ1
ð32Þ
ðs þ 2ÞX ðsÞ þ 2ðs þ 1ÞY ðsÞ ¼ 1
ð33Þ
ðs þ 1ÞX ðsÞ þ sY ðsÞ ¼ and
Multiplying (32) by 2ðs þ 1Þ and (33) by s and then
subtracting, we get
½2ðs þ 1Þ2 sðs þ 2ÞX ðsÞ ¼ 2 s
Applications of Laplace Transform
ðs þ 2s þ 2ÞX ðsÞ ¼ ðs þ 2Þ
2
or
sþ2
sþ2
¼
þ 2s þ 2
ðs þ 1Þ2 þ 1
"
#
sþ1
1
¼
þ
:
ðs þ 1Þ2 þ 1 ðs þ 1Þ2 þ 1
s2
ð34Þ
1
s2
¼ 2
: ð36Þ
þ1 s þ1
Multiplying (35) by s and adding to (36), we get
sX ðsÞ þ s2 Y ðsÞ ¼ 1 Taking inverse Laplace transform, we get
s3 X ðsÞ þ sX ðsÞ ¼
xðtÞ ¼ et cos t et sin t:
Further, putting the value of X ðsÞ in (33) ,we get
ðs þ 2Þðs þ 2Þ
2ðs þ 1ÞY ðsÞ ¼ 1 þ 2
s þ 2s þ 2
s2 þ 2s þ 2 þ s2 þ 4s þ 4
¼
s2 þ 2s þ 2
2
2s þ 6s þ 6
¼ 2
s þ 2s þ 2
or
s2 þ 3s þ 3
Y ðsÞ ¼
ðs þ 1Þðs2 þ 2s þ 2Þ
1
1
¼
þ
s þ 1 s2 þ 2s þ 2
1
1
þ
¼
:
s þ 1 ðs þ 1Þ2 þ 1
Taking inverse Laplace transform, we have
yðtÞ ¼ e
t
7.35
Using the given condition, the above equations
transform to
s
s2 X ðsÞ sY ðsÞ ¼ 2
ð35Þ
s þ1
and
or
X ðsÞ ¼ n
t
e sin t:
s2
s2
s2
2s2
þ
¼
:
s2 þ 1 s2 þ 1 s2 þ 1
Thus
X ðsÞ ¼
2s2
2s
¼
:
ðs2 þ 1Þðs3 þ sÞ ðs2 þ 1Þ2
Taking inverse Laplace transform, we have
xðtÞ ¼ t sin t:
Putting the value of X ðsÞ in (35), we get
2s3
ðs2
þ 1Þ
2
sy ðsÞ ¼
s
s2 þ 1
or
yðsÞ ¼
s2 1
ðs2
þ 1Þ
2
¼
s2
1
2
:
2
þ 1 ð s þ 1Þ 2
Taking inverse Laplace transform, we get
yðtÞ ¼ sin t ðsin t t cos tÞ ¼ t cos t:
Hence the required solution is
xðtÞ ¼ t sin t and yðtÞ ¼ t cos t:
EXAMPLE 7.70
Use Laplace transform method to solve the simultaneous equations: D2 x Dy ¼ cos t; Dx þ D2 y ¼
sin t; given that x ¼ 0; Dx ¼ 0; y ¼ 0 Dy ¼ 1,
when t ¼ 0.
EXAMPLE 7.71
Using Laplace transform, solve the integral equation
Z t
t
yðt uÞ sin u du:
yðtÞ ¼ 1 e þ
Solution. Taking Laplace transforms of the given
equations, we have
s
s2 X ðsÞ sxð0Þ x0 ð0Þ sy ðsÞ yð0Þ ¼ 2
s þ1
and
Solution. Taking Laplace transform of both sides of
the given integral equation, we have
s X ðsÞ xð0Þ þ s2 yðsÞ syð0Þ y0 ð0Þ
¼
1
:
s2 þ 1
0
Lfyg ¼ Lf1g Lfet g þ Lfyg: Lfsin ug
1
1
1
:
þ Lfyg: 2
¼ s sþ1
s þ1
This relation yields
1
1
1
¼ Lfyg 1 2
s þ1
s sþ1
7.36
or
n
Engineering Mathematics-II
(i) ddt2x þ 2b dtdx2 + l2x = 0, x(0) = x0 (0) = 0.
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
Ans. x(t) = ebt ðc1 sin l2 b2 t þ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
c2 cos l2 b2 t
2
s2 þ 1 1
1
s2 þ 1
Lfyg ¼
¼
s2
s sþ1
s3 ðs þ 1Þ
2
2
1
1
¼ þ
s s þ 1 s2 s3
(by partial fractions).
2. Solve y0 2ty = 0, y(0) = 1 and show that its
solution does not have Laplace transform.
2
Ans. y(t) = et (not of exponential order)
3. Solve ty 00 + y = 0, y(0) = 0.
Hint: Proceed as in Example 7.20
1
X
ð1Þn tnþ1
:
Ans. C
ðn þ 1Þ ! n !
n¼0
4. Given that I = Q = 0 at t = 0, find current I in the
LC circuit given for t > 0 in Figure 7.15.
Taking inverse Laplace transform, we get
1
yðtÞ ¼ 2 2 et t þ t2 :
2
EXERCISES
1. Solve the following initial value problems:
(a) y0 (t) + 3y(t) = 0, x (1) = 1.
Ans. y(t) = e3ð1tÞ
L
2
(b) ddt2y +y = 1, y(0) = y (0) = 0.
Ans. 1 cos t
E
C
(c) y00 + y =et , y(0) = A, y0 (0) = B.
Ans. y(t) = 12 et + A 12 cos t + B þ 12 sin t
Figure 7.15
2
(d) ddt2y + y = 0, y(0) = 1, y0 (0) = 0.
Ans. y(t) = cos t.
2
(e) ddt2y + a2y = f (t), y(0) = 1, y0 (0) = 2.
Hint: Y(s) =
theorem
L1
s2
s2 þa2
FðsÞ
s 2 þ a2
0
þ sFðsÞ
2 þa2 . But by Convolution
¼ f ðtÞ
sin at
and so
a
2
L1 2
s þ a2
s
yðtÞ ¼ L
2
s þ a2
sin at
þ f ðtÞ
a t
R
Ans. cos at 2 sina at þ 1a f ðuÞ sin aðt uÞ du
1
0
2
Hint: The differential equation governing the
Rt
1
circuit is LdI
dt þ C IðuÞ du = E. Application of
(f) ddt2y + y = 3 sin 2t, y(0) = 3, y0 (0) = 1.
Laplace transform yields
E
EC
Ls F(s) + FðsÞ
¼ E
2
Cs ¼ s , that is, F(s) =q
LCs
ffiffiffi þ1 Lðs2 þLC1 Þ
ffi t:
Ans. I(t) = E CL sin p1ffiffiffiffi
LC
5. Given that I = Q = 0 at t = 0, find charge and
current in the circuit shown in Figure 7.16.
2 henry
.02 farad
Ans. sin 2t + 3 cos t + 3 sin t
t
(g) ddt2x þ 5 dx
(t
dt þ 6x ¼ 2e
0
x (0) = 0
2
0), x(0) = 1 and
Ans. et þ e2t e3t , t
(h)
2
d x
dt2
þ
6 dx
dt
0
+ 9x = 0, x(0) = x (0) = 0.
Ans. x(t) = 0
300V
16 ohms
Figure 7.16
Hint: The governing equation is
d2 Q
dt2
150
þ 8 dQ
dt þ 25Q ¼ 150, F(s) = sðs2 þ8sþ25Þ and
so inversion gives Q(t) = 6 6e4t cos 3t 8e4t sin 3t. Then I(t) = 50 e4t sin 3t.
Applications of Laplace Transform
6. Solve the following systems of differential
equations:
(a) dx
dy + x y = 1 + sin t ,
with x(0) = 0, y(0) = 1
dy
dt
dx
dt + y = t sin t,
dz
dt
(d) f (t) =
7. Solve y 00 + 4y = 4 cos 2t, y(0) = y0 (0) = 0. Does
resonance occur in this case?
Hint: Y(s) =
4s
t
and so y(t) = 4 4 sin 2t = t sin 2t.
ðs2 þ4Þ2
Note that y(t) ? ! 1? as t ! 1??. Hence,
there shall be resonance.
8. Solve the following difference equations:
(a) 3y(t) 4y(t 1) + y(t 2) = t, y(t) = 0 for t <
0.
½t P
Ans. y(t) = 3t þ 12
1 31n ðt nÞ
n¼1
(b) an+2 2an+1 + an = 0, a0 = 0, a1 = 1.
Ans. an = n
(c) an = an1 + 2an2, a0 = 1, a1 = 8.
0
(d) an = 2an1 an2, a1 = 1.5, a2 = 3
Ans. 1.5 n
(e) y (t) y(t 1) = t
2
Ans. y(t) = 2
½t
P
ðtnÞnþ3
n¼0
ðnþ3Þ !
(f) y (t) y(t 1) = (t), y(t) = y (t) = 0, t 0.
Hint: s2L{y(t)} es L{y(t)} = L{ (t)} and so
L{y(t)} = 2 1 es . But
s 1 2
(s
ðtnÞ2nþ1
ns
L1 se2nþ2 = ð2nþ1Þ ! for t n
0
otherwise:
½t
P ðtnÞ2nþ1
Hence y(t) =
ð2nþ1Þ!
n¼0
0
Ans. f (t) = 1+p2ffiffi3sin
Ans. y(t) = 12(sinh t + sin t)
sin uðt uÞ du
0
(e) y (t) + 3y(t) + 2
Rt
Ans. 0
yðuÞdu = t, y(0) = 1.
0
s2 þ1
sðs2 þ3sþ2Þ.
Use partial fractions
Hint: L{y(t)} =
and then use inversion to give y(t) =
1
t
5 2t
2 2e þ 2 e
@u
10. Solve the partial differential equation x @u
@t þ @x
= x, x > 0, t > 0 subject to the conditions u(x, 0)
= 0 for x > 0 and u(0, t) = 0 for t > 0.
Hint: Using Laplace transform,
sU(x, s)h =
s) =
1
s2
d
dxU(x,
s) + x
1 2
2x
x
s,
integrating
factor is e and so U(x,
i
12sx2
1e
. Inversion yields u(x, t) ¼
for t < x2 =2
.
for 2t > x2
t
x2 =2
@ u
11. Find the bounded solution of @u
@t ¼ @x2 , x > 0,
t > 0 for u(0, t) = 1, u(x, 0) = 0.
2
Ans. an = 3(2n ) 2ð1Þn , n
9. Solve the integral equations:
Rt
(a) f (t) = 1 + cosðt uÞf (u) du
Rt
0
Ans. y(t) = cos t , z(t) = sin t
00
Ans. y(t) = tet
Rt
(c) y(t) = t + 16 yðuÞ ðt uÞ3 du
0
= y with y(0) = 1, z(0) = 0.
yðuÞ cos (t u) du
0
Ans. x(t) = t + sin t, y(t) = t + cos t
(b) dy
dt = z ,
Rt
(b) y(t) = sin t + 2
7.37
n
pffiffi 3
t=2
2 t e
Hint: Application
of Laplace
transform yields
pffi
pffi
U(x, s) = c1 e s x þ c2 e s x for bounded u(x, t),
U(x, s) must be bounded and so c1 = 0. Further
U(0, s) = L{u(0, t)} = L{1} = 1/s.pTherefore,
ffi
1/s = c2 and so U(x, s) = erfc (x/2 t).
12. Solve @@t2u = a2 @@xy2 , x > 0, t > 0 for u(x, 0) = 0,
yt(x, 0) = 0, x > 0, y(0, t) = f (t) with f (0) = 0
and lim y(x, t) = 0.
2
2
x!1
Hint: see Example 7.54 Ans. y(x, t) = f t ax H t ax
13. Solve @@t2u ¼ @@xu2 , x>0, t>0 for u(0, t) = 10 sin 2t,
u(x, 0) = 0, ut(x, 0) = 0 and lim u(x, t) = 0
2
2
x!1
Ans. uðx; tÞ ¼
10 sin 2ðt xÞ
0
for t > x
for t < x:
14. Evaluate the integrals:
R1
(a) J0 ðtÞ dt
0
Ans. 1
7.38
n
Z1
ðbÞ
Engineering Mathematics-II
et e2t
dt
t
0
1
1
Hint: L{et e2t g = sþ1
sþ2
and so
n t 2t o R1 1
1
sþ2
¼
dt
¼
log
L e e
t
tþ1
tþ2
sþ1 ;
s
that is,
R1 st et e2t sþ2
e
dt ¼ log sþ1
. Taking s = 0,
t
0
we get the value of the given integral equal to
log 2
R1
(c) cos x2 dx (Proceed as in Example 7.59)
0
pffiffi
Ans. 12 2
R1
at
(d) xx2sin
þa2 dx; a, t > 0.
0
Ans. f (t) = 2 et
ENGINEERING MATHEMATICS—YEAR 2007
SEMESTER—2
GROUP—A
(Multiple Choice Type Questions)
1. Choose the correct alternatives for any ten of the following questions:
(i) Laplace transform of the function cos (at) is
s
s − a2
s
(c) 2
s + a2
(a)
Ans. (c)
2
(b)
(d)
a
s + a2
2
1
s − a2
2
s
s + a2
2
⎡1 0 1⎤
(ii) The rank of the matrix A = ⎢⎢1 1 1⎥⎥ is
⎢⎣1 0 1⎥⎦
(a) 0
(b) 1
(c) 3
(d) 2
Ans. (d) 2
2
d 2 y ⎛ dy ⎞
(iii) The order and degree of the differential equation
+ ⎜ ⎟ = y are
dx 2 ⎝ dx ⎠
(a) 2, 2
(b) 2, 1
(c) 1, 2
Ans. (a) 2, 2
(d) 1, 1
(a) 0
(c) – abc
Ans. (a) 0
(b) abc
(d) 2 abc
0 − a −b
c is
(iv) Value of the determinant a 0
b −c 0
(v) Integrating factor of
(a) e x
(c) x
dy
+ y = 1 is
dx
(b) x2
(d) 2
dx
Ans. (a) e x Hint: e ∫ = e x
(vi) The operator equivalent to shift operator E is
(a) I + Δ
(c) I − Δ
Ans. (a) I + Δ
(b) ( I + Δ )
(d) I − Δ 2
−1
10 ¥ 1 = 10
Q.2

Engineering Mathematics-II
(vii) The number of significant digits in 3·0044 is
(a) 5
(b) 2
(c) 3
(d) 4
Ans. (a) 5
0
(viii) If a, b are the roots of the equation x − 3 x + 2 = 0, then b
3
1
(b)
(a) 6
2
(c) –6
(d) 3
Ans. (a) 5 Hint: ( x − 1)( x − 2) = 0, α = 1, β = 1.
2
a b
0 0 is
−a a
⎡ 1 2 3⎤
(ix) The sum of the eigen values of A = ⎢⎢0 2 3⎥⎥ is
⎢⎣0 0 2⎥⎦
(a) 5
(b) 2
(c) 1
(d) 6
Ans. (a) 5
(x) ( Δ − ∇) x 2 is equal to
(b) – 2h2
(a) h2
2
(c) 2h
(d) None of these.
Ans. (c) 2h2
{
Hint: ( Δ − ∇) x 2 = Δx 2 − ∇x 2 = ( x + h) − x 2 − x 2 − ( x − h)
2
2
} = 2h
2
(xi) If a linear transformation T : IR2 → IR2 be defined by T ( x1 , x2 ) = ( x1 + x2 , 0 ) then ker (T) is
(b) {(1, 0 )}
(a) {(1, −1)}
{(0, 0)}
{(1, −1)}
(c)
Ans. (a)
(d)
{(1, 0) , (0,1)}
2000 2001 2002
(xii) The value of 2003 2004 2005 is
2006 2007 2008
(a) 2000
(b) 0
(c) 45
(d) None of these.
Ans. (b)
C −C
2
1→
Hint: ⎯⎯⎯⎯⎯
C −C
3
2
2000 1 1
2003 1 1 = 0
2004 1 1
(xiii) The value of K for which the vectors (1, 2, 1), (K, 1, 1) and (0, 1, 1) are linearly dependent
(a) 1
(c) 0
Ans. (c)
(b) 2
(d) 3.
Solved Question Papers
1 2 1
Hint: k 1 1 = 0 or k = 0
0 1 1
GROUP—B
(Short Answer Type Questions)
Answer any three of the following
2. Apply convolution theorem to find the inverse of
Ans. Same as 5(a) of 2005.
b2 + c2
3. Prove that: b 2
c2
a2
c 2 + a2
c2
b2 + c2
Ans. We have, Δ = b 2
c2
a2
c 2 + a2
c2
+9
)
2
a2
b2
a2 + b2
−2b 2
b 2 R1 = R1 − ( R2 + R3 )
2
a + b2
−c 2
−b2
c2 + a2
b2
1
1
a2 + b2
c2
0
= 2 × b2 × c2 1
1
−c 2
0
(s
a2
b2 = 4a2 b2 c 2
a2 + b2
−2c 2
c2 + a2
c2
0
= b2
c2
s
2
a2 + b2
R′ 2 = R2 − R3
c2
a2 + b2
1
1
c2
⎧⎪
⎛ c2 + a2 ⎞
a 2 + b 2 ⎞ ⎪⎫
2 ⎛
= −2b2 c 2 ⎨ −b2 ⎜
−
+
−
1
c
1
⎬
⎟⎠
⎜⎝
c 2 ⎟⎠ ⎪⎭
⎝ b2
⎪⎩
= 2b 2 c 2 0
c2 + a2
b2
−b2
{(
1−
) (
= −2b2 c 2 − c 2 + a2 − b2 + c 2 − a2 − b2
)}
= 4a b c
2
2 2
4. Solve the differential equation by Laplace transform :
d2x
+ 4 x = sin 3t , x(0) = 0, x ′(0) = 0
dt 2

Q.3
Q.4

Engineering Mathematics-II
d2x
+ 4 x = sin 3t
dt 2
Taking Laplace transform on both side
Ans. We have
s 2 X ( s) − sx(0) − x(0) + 4 X ( s) =
or,
s 2 X ( s) − s.0 − 0 + 4 X (0) =
or, ( s 2 + 4) X ( s) =
∴ X ( s) =
3
s +9
3
s + 32
2
3
s +9
2
2
3
3⎛ 1
1 ⎞
= ⎜
−
⎟
( s 2 + 9)( s 2 + 4) 5 ⎝ s 2 + 4 s 2 + 9 ⎠
∴ x(t ) = L−1 { X ( s)}
3
3 sin 3t
=
sin 2t − ⋅
5⋅ 2
5 3
3
1
= sin 2t − sin 3t
10
5
5. Solve ( x 2 + y 2 + 2 x ) dx + xydy = 0
Ans. We have ( x 2 + y 2 + 2 x ) dx + xydy = 0
M = x2 + y2 + 2x
N = xy
∂M
∂N
∴
= 2y
=y
∂y
∂x
∂M ∂N
∴
≠
∂y
∂x
here
∴ Given equation is not exact.
∂M ∂N
−
∂y
∂x 2 y − y 1
Now,
=
=
N
xy
x
1
So, I. F. = e ∫ x
dx
= e log x = x
Multiplying both side of given equation by x
( x 2 + xy 2 + 2 x 2 ) dx + x 2 ydy = 0
or,
( x 3 + 2 x 2 ) dx + xy 2 dx + x 2 ydy = 0
(x
)
(
)
1
+ 2 x 2 dx + d x 2 y 2 = 0
2
Integrating both side
or,
3
x 4 2 x3 1 2 2
+
+ x y =c
4
3
2
c — integrating constant.
Solved Question Papers

Q.5
6. Show that W = {x, y, z ) ∈ R3 , 2 x − y + 3 z = 0} is a subspace of R3. Find basic of W. What is dimension?
Ans. Same as 13(b) of 2008.
⎡1 0 2⎤
⎢
⎥
7. If A = 0 −1 1 then verify that A satisfies its own characteristic equation. Hence find A−1.
⎢
⎥
⎢⎣0 1 0 ⎥⎦
Ans. Same as 8(b) of 2008.
GROUP—C
(Long-Answer Questions)
Answer any three questions
⎛ Δ2 ⎞
8. (a) Evaluate : ⎜ ⎟ x 3
⎝ E⎠
⎛ Δ2 ⎞
( E − 1) 3
x
Ans. ⎜ ⎟ x 3 =
E
⎝ E⎠
2
⎛ E 2 − 2 E + 1⎞ 3
=⎜
⎟⎠ x
E
⎝
= ( E − 2 + E − 1) x 3
= ( x + h ) 3 − 2 x 3 ( x − h) 3
= x 3 + 3 x 2 h + 3 xh2 + h3 − 2 x 3 + x 3 − 3 x 2 h
+3 xh2 − h3 = 6 xh2
(b) Find the missing data in the following table
x
–2
–1
0
1
2
f (x)
6
0
–
0
6
Ans. As y has four given values
Δ 4 y0 = 0
∴ ( E − 1) 4 y0 = 0
or, ( E 4 − 4 E 3 + 6 E 2 − 4 E + 1) y0 = 0
or,
y4 − 3 y3 + 6 y2 − 4 y1 − y0 = 0
− y4 + 3 y3 + 4 y1 − y0
6
−6 + 3.0 + 4.0 − 6
=
= −2
6
∴ y2 =
(c) Show that, (3, 1, –2), (2, 1, 4) and (1, –1, 2) form a basis of R3.
Ans. Consult 8(a) of 2005.
Q.6

Engineering Mathematics-II
∞
sin t
π
dt = , by Laplace transform.
t
2
0
9. (a) Prove that ∫
Ans. See 11(b) of 2005
(b) Apply convolution theorem to prove that
t
t
∫ sin u cos(t − u)du = 2 sin t
0
t
Ans. Let, f (t ) = ∫ sin u cos(t − u)du
0
Using convolution theorem,
L { f (t )} = F1 ( s) ⋅ F2 ( s)
= α {sin t } ⋅ α {cos t }
1
s
= 2
⋅ 2
s +1 s +1
1 2s
s
= 2
= ⋅ 2
s +1 2 s +1
1
d ⎛ 1 ⎞
= ( −1) ⋅ ⎜ 2 ⎟
2
ds ⎝ s + 1⎠
f (t ) =
1 1
1
⋅ t ⋅ sin t = sin t (Proved)
2
2
(c) Solve
dy tan y
−
= (1 + x )e x sec y
dx 1 + x
∴
Ans. We have dy − tan y = (1 + x )e x sec y
dx 1 + x
or,
cos y
dy
tan y
− cos y
= (1 + x )e x
dx
1+ x
or,
cos y
dy
tan y
− cos y
= (1 + x )e x
dx
1+ x
Let, sin y = z
dy dz
=
dx dx
dz
1
−
z = (1 + x )e x ..............(i)
dx 1 + x
∴ cos y
or,
⎛ 1 ⎞
So, I. F. = e
=
∫ ⎝⎜ 1+ x ⎠⎟ dx
1
1+ x
Multiplying both sides of (i) by
1
1+ x
Solved Question Papers
dz 1
1
⋅
−
⋅ z = ex
dx 1 + x (1 + x )2
⎛ z ⎞
∴ d⎜
= e x dx
⎝ 1 + x ⎟⎠
Integrating both sides
z
= e x + c, c — integrating constant
1+ x
or, sin y = e x + c (1 + x ) [as z = sin y]
(
)
10. (a) Solve by Cramer’s rule :
x+ y+z = 7
x + 2 y + 3 z = 10
x− y+z =3
Ans. Same as 14(b) of 2008.
(b) Find general solution of p = cos ( y − px ) , where p =
Ans. We have, p = cos ( y − px )
y − px = cos −1 p
or,
Differentiating both side w.r.t. x
dy
dp
−1
dp
− p.1 − x
=
⋅
2 dx
dx
dx
1− p
dp
1
dp
=−
2 dx
dx
1− p
⎛
⎞
1
dp
=0
or, ⎜ x −
⎟
2
1 − p ⎠ dx
⎝
or,
p− p−x
when,
or,
x−
x=
1
1 − p2
=0
1
1 − p2
1
1 − p2
x2 − 1
∴ p2 =
x2
x2 − 1
∴p=
x
∴ Singular solution
or,
x2 =
y=
x 2 − 1 + cos −1
x2 − 1
x
dy
dx

Q.7
Q.8

Engineering Mathematics-II
again when
dp
= 0 or, p = c, c → constant, general solution is y = e x + cos −1 c
dx
dy y log y y (log y )
+
=
dx
x
x2
2
(c)
Solve:
dy y
y
dy y
y
+ log y = 2 (log y )2
+ log y = 2 (log y )2 dividing both sides by
dx x
x
dx x
x
Ans. We have
1
y (log y )
2
1
log y
Assume, z =
∴
dy 1 1
1
+
= 2 ..........(i)
dx x log y x
dz
1
dy
=−
dx
y(log y )2 dx
equation (i) become
dz z
1
− + = 2
dx x x
dz z
1
or,
− = − 2 ............(ii)
dx x
x
∴ I. F. = e
−
∫
dx
x
=
1
x
multiplying both side of (ii) by
1 dz z
1
− 2 =− 3
x dx x
x
or,
d ⎛ z⎞
1
⎜ ⎟=− 3
dx ⎝ x ⎠
x
Integrating,
or,
1
x
1
⎛ z⎞
d ⎜ ⎟ = − 3 dx
⎝ x⎠
x
z
1
= + 2 + c, c → Constant
x
2x
or,
1
1
= + 2 +c
x log y
2x
or,
1
1
1 + 2cx 2
=
+ cx =
log y 2 x
2x
∴ log y =
2x
1 + 2cx 2
2x
or,
y = e1+ 2 cx
2
11. (a) Solve: ( D 2 − 2 D ) y = e x sin x, where D ≡
d
dx
Solved Question Papers
Ans. We have ( D 2 − 2 D ) y = e x sin x
The auxiliary equation is m2 − 2m = 0
or, m( m − 2) = 0
∴ m = 0, 2
∴ C.F. = ( A + Bx )e2 x, A and B constants
1
(e x sin x )
D2 − 2D
1
= ex
⋅ sin x
2
( D + 1) − 2 ( D + 1)
Again, P. I. =
1
1
(sin x )
⋅ sin x = e x ⋅ 2
D2 + 2D + 1 − 2D − 1
D −1
1
1
sin x = − e x sin x
= ex ⋅
2
−1 − 1
= ex ⋅
∴ Total solution is
1
y = ( A + Bx )e 2 x − e x sin x
2
dx
dy
− 7 x + y = 0 and
(b) Solve
− 2x − 5 y = 0
dt
dt
dx
Ans. We have
− 7x + y = 0
dt
dy
− 2x − 5 y = 0
dt
d
Let, D ≡
dt
∴ ( D − 7) x + y = 0.........(i)
( D − 5) y − 2 x = 0...........(ii)
(i) × 2 + (ii) × (D – 7)
2( D − 7) x + 2 y = 0
+( D − 7) ( D − 5) y − 2 x( D − 7) = 0
( D − 7) ( D − 5) y + 2 y = 0
or,
( D 2 − 12 D + 35) y + 2 y = 0
or,
( D 2 − 12 D + 37) y = 0
Auxiliary equation is m2 − 12m + 37 = 0
∴m =
12 ±
(12)2 − 4.1.37
2
∴ Solution is y = e 6i ( A cos t + B sin t ) A and B constant from (ii)
= 6±i

Q.9
Q.10

Engineering Mathematics-II
1
1
( D − 5) y = ( Dy − 5 y )
2
2
1
x = e 6t [ A( − sin t ) + B cos t ] + 6e 6t ( A cos t + B sin t ) − 5e et ( A cos t + B sin t )
2
1
= e 6t [ − A + 6 B − 5 B ] sin t + e 6t [ B + 6 A − 5 A] cos t
2
1 6t
= e {( B − A) sin t + ( A + B) cos t }
2
= e 6t {C cos t + D sin t }
x=
or,
{
}
{
A+ B
;
2
∴ Required solution is
where, C =
}
D=
B−A
2
x = e 6t {C cos t + D sin t }
y = e 6t { A cos t + B sin t }
(c) Solve by Gauss elimination method
2 x + 2 y + z = 12
3x + 2 y + 2 z = 8
5 x + 10 y − 8 z = 10
Ans. We have,
2 x + 2 y + z = 12...........(i)
3 x + 2 y + 2 z = 8...........(ii)
5 x + 10 y − 8 z = 10........(iii)
⎛ 3⎞
(i) × ⎜ − ⎟ + (ii) gives
⎝ 5⎠
−4 y +
34
z = 2...........(iv)
5
⎛ 2⎞
(i) × ⎜ − ⎟ + (iii) gives
⎝ 5⎠
21
−2 y + z = 8............(v)
5
⎛ 1⎞
Again (iv) × ⎜ − ⎟ + (v) gives
⎝ 2⎠
4
z=7
5
35
∴z =
4
34 35
from (iv) − 4 y + ⋅ = 2
5 4
Solved Question Papers
or,
y=

Q.11
115
8
115
35
− 8 ⋅ = 10
8
4
575 − 280
= 10
or, 5 x +
4
295
∴ 5x +
= 10
4
295 40 − 295
or, 5 x + 10 −
=
4
4
40 − 295
51
∴x=
=−
20
4
from (i) 5 x + 10 ⋅
12. (a) Use Lagrange’s interpolation formula to find f (x), where f (0) = –18, f (1) = 0, f (3) = 0, f (5) = –248,
f (6) = 0 and f (9) = 13104
Ans. We have,
0 1 3
5
6
9
x:
f ( x ) −18 0 0 −248 0 13104
Now using Lagrange’s interpolation,
( x − 1)( x − 3)( x − 5)( x − 6)( x − 9)
× ( −18)
(0 − 1)(0 − 3)(0 − 5)(0 − 6)(0 − 9)
( x − 0)( x − 1)( x − 3)( x − 6)( x − 9)
+
× ( −248)
(5 − 0)(5 − 1)(5 − 3)(5 − 6)(5 − 9)
( x − 0)( x − 1)( x − 3)( x − 5)( x − 6)
+
× (13104 )
(9 − 0)(9 − 1)(9 − 3)(9 − 6)(9 − 9)
⎧ ( x − 5)( x − 9) x ( x − 9) × 31 x ( x − 5) 91⎫
= ( x − 1)( x − 3)( x − 6) ⎨
−
+
⎬
f ( x) =
45
⎩
20
36
⎭
⎛ 180 x + 180 x + 180 ⎞
= x 3 − 10 x 2 + 27 x − 18 ⎜
⎟⎠
180
⎝
(
2
)
= x 5 − 9 x 4 + 18 x 3 − x 2 + 9 x − 18
(b) Apply appropriate interporlation formula to calculate f(2.1). Correct upto two significant figures from
the following data.
x
0
2
4
6
8
10
f (x)
1
5
17
37
45
51
Q.12

Engineering Mathematics-II
Ans. From the given data we can design the following difference table:
x
0
2
4
6
8
10
f ( x ) Δf ( x ) Δ 2 f ( x ) Δ 3 f ( x ) Δ 4 f ( x ) Δ 5 f ( x )
1
5
4
8
0
−20
−10
17
12
8
−20
−30
37
20
−12
+10
45
8
−2
51
6
x − x0 2.1 − 0
=
= 1⋅ 05
2
2
Using Newton’s forward interpolation
Ans. Here, x 0 = 0, h = 2, d =
∴
d ( d − 1) 2
Δ f ( x0 )
2!
d ( d − 1)( d − 2) 3
d ( d − 1)( d − 2)( d − 3) 4
+
Δ f ( x0 ) +
Δ f ( x0 )
3!
4!
f ( x ) = f ( x0 ) + d Δf ( x0 ) +
1⋅ 05(1⋅ 05 − 1)
⋅8
2
1⋅ 05(1⋅ 05 − 1)(1⋅ 05 − 2)
(0)
+
6
1⋅ 05(1⋅ 05 − 1)(1⋅ 05 − 2)(1⋅ 05 − 3)
( −20)
+
24
1×05(1×05-1)(1×05-2)(1×05-3)(1×05-4)
(50)
+
120
= 5 ⋅ 21.
= 1 + (1⋅ 05) ⋅ 4 +
13. (a) Apply the method variation of parameters to solve the equation
Ans. We have
d2 y
+ y = sec3 x.tan x
dx 2
Auxiliary equation is m2 + 1 = 0 ∴ m = ±i
∴ P. F. = A cos x + B sin x
∴W =
cos x sin x
= cos 2 x + sin 2 x = 1
− tan x cos x
∴ P. I. = sin x ∫ cos x sec3 x tan x dx − cos x ∫ sin x sec3 x tan x dx
= sin x ∫ sec3 x tan x dx − cos x ∫ sec2 x tan 2 x tan x dx
= sin x ∫ tan x sec2 xdx − cos x ∫ tan 2 x sec2 x dx
d2 y
+ y = sec3 x. tan x
dx 2
∫
∫
= sin x ∫ tan x d (tan x ) − cos x ∫ tan
= sin x
(tan x )2
− cos x
(tan x )3
2
∴ Required solution is
3
=
Solved Question Papers
2

x d (tan x )
1
1
sin x tan 2 x − cos x tan 3 x
2
3
1
1
y = A cos x + B sin x + sin x ⋅ tan 2 x − cos x tan 3 x
2
3
(b) Expand by Laplace’s method
0
a
−a 0
−b − d
−c −e
b
d
0
−f
c
e
= ( af − be + cd )2
f
0
Ans. We have
0
a
−a 0
Δ=
−b − d
−c −e
b
d
0
−f
c
e
f
0
Expanding using Laplace method
Δ = ( −1)
1+ 2 +1+ 2
0 a 0
−a 0 − f
+ ( −1)
1+ 2 +1+ 4
0 c −d
− a 0 −e
f
b −d
1+ 2 +1+ 3 0
+ ( −1)
0
− a d −c
f
0
0
b −b
1+ 2 + 2 + 3 a
+ ( −1)
−f
0 d −c
f
0
a c −b 0
c −b f
1+ 2 + 3 + 4 b
+ ( −1)
0 e −c − f
d e −c −e
= (0 + a 2 ) (0 + f 2 ) − (0 + ab) (0 + ef ) + (0 + ca) ( df − 0)
+ ( ad − 0) + (0 + cf ) + (be − cd )(be − cd )
+ ( −1)
1+ 2 + 2 + 4
= a 2 f 2 − 2af (be − cd ) + (be − cd )2 = ( af − be + cd )2
⎧ sin t ⎫
⎧ sin at ⎫
−1 1
, find ⎨
(c) Given that L ⎨
⎬ = tan
⎬
t
s
⎩
⎭
⎩ t ⎭
Ans. According to change of scale property, we know if L { f (t )} = F ( s), then L { f ( at )} =
1 ⎛ s⎞
F⎜ ⎟
a ⎝ a⎠
⎛ ⎞
1 ⎛ s⎞
⎡ sin at ⎤
⎡ sin at ⎤
⎛ s⎞
−1 ⎜ 1 ⎟
∴ L⎢
= a. L ⎢
= a ⋅ F ⎜ ⎟ = tan ⎜ ⎟ = cot −1 ⎜ ⎟
⎥
⎥
s
⎝ a⎠
a ⎝ a⎠
⎣ t ⎦
⎣ at ⎦
⎜⎝ ⎟⎠
a
1
14. (a) Assuring the orthogonal properties of Legendre function prove that,
∫ [ p ( x)]
2
n
−1
dx =
2
2n + 1
Q.13
Q.14

Engineering Mathematics-II
Ans. The Legendre’s equation can be written as
d ⎧
2 dy ⎫
⎨(1 − x ) ⎬ + n( n + 1) y = 0
dx ⎩
dx ⎭
As, pn ( x ) is the solution,
d ⎧
2 dpn ⎫
⎨(1 − x )
⎬ + n( n + 1) pn = 0
dx ⎩
dx ⎭
Again if pm ( x ) is a solution of
d ⎧
2 dpm ⎫
⎨(1 − x )
⎬ + n( n + 1) pm = 0
dx ⎩
dx ⎭
Doing (i) × pm − (ii) × pn
pm
d ⎧
d ⎧
2 dpn ⎫
2 dpm ⎫
⎨(1 − x )
⎬ − pn
⎨(1 − x )
⎬
dx ⎩
dx ⎭
dx ⎩
dx ⎭
+ pm pn {n( n + 1) − m( m + 1)} = 0
Integrating both side w.r.t x from –1 to 1, we get
1
1
1
1
dpn
⎡
2 dpn ⎤
2 dpn
⎢ pm (1 − x ) dx ⎥ − ∫ dx (1 − x ) dx dx
⎣
⎦ −1 −1
dpn
⎡
2 dpm ⎤
2 dpm
⎢ pn (1 − x ) dx ⎥ − ∫ dx (1 − x ) dx dx
⎣
⎦ −1 −1
1
+( n − m)( n + m + 1) ∫ pm pn dx = 0
−1
1
∴ ∫ pm ( x ) pn dx = 0
−1
But (1 − 2 xh + h2 )
−
1
2
∝
= ∑ hn pn ( x )
n= 0
∝
or,
(1 − 2 xh + h2 ) −1 = ∑ h2 n [ pn ( x ) ]
2
n= 0
∝
∝
+2∑ ∑ hm + n pm ( x ) pn ( x )
[squaring both side]
n= 0 m= 0
Integrating both side w.r.t. x and taking limit –1 to 1, we get
∝
1
1
2
⎡⎣log(1 − 2 xh + h2 ) ⎤⎦ = ∑ h2 n ∫ [ pn ( x ) ] dx
−1
2h
n= 0
−1
1
−
or,
1
⎡ h2 h 4
⎤ ∝
h2 n
2
2 ⎢1 + + + ... +
+ ...⎥ = ∑ h2 n ∫ [ pn ( x ) ] dx
3
5
2n + 1
⎣
⎦ n = 0 −1
Solved Question Papers

Q.15
equating the coefficient of h2n
1
∫ [ p ( x)]
2
n
dx =
−1
2
(Proved)
2x + 1
14. (b) Show that J 1 ( x ) =
2
2
πx
⋅ sin x
( )
n
1
⎛ x⎞ ⎧ 1
x
−
Ans. As J n ( x ) = ⎜ ⎟ ⎨
⎝ 2 ⎠ ⎩ Γ ( x + 1) 1! Γ ( x + 2) 2
2
4
6
⎫⎪
1
1
⎛ x⎞
⎛ x⎞
+
⎜⎝ ⎟⎠ −
⎜⎝ ⎟⎠ + ...⎬
2! Γ ( x + 3) 2
3! Γ ( x + 4) 2
⎭⎪
Putting n =
1
2
⎧
1
2
4
1
1
1
1 1
⎛ x ⎞ 2 ⎪⎪
⎛ x⎞
⎛ x⎞
−
+
J 1 ( x) = ⎜ ⎟ ⎨
⎜⎝ ⎟⎠
⎜⎝ ⎟⎠ −
⎝
⎠
1
1
1
2
2
2
3!
⎛
⎞
⎛
⎞
⎛
⎞
⎛ x⎞
2
⎪ Γ ⎜1 + ⎟ 1! Γ ⎜ 2 + ⎟
2! Γ ⎜ 3 + ⎟
Γ⎜ ⎟
⎝
⎠
⎝
⎠
⎝
⎠
⎝ 2⎠
2
2
2
⎩⎪
⎫
6
⎪⎪
⎛ x⎞
⋅ ⎜ ⎟ + ...⎬
⎝ 2⎠
⎪
⎭⎪
⎧
2
4
⎪
1
1
1
⎛ x⎞ ⎪ 1
⎛ x⎞
⎛ x⎞
=⎜ ⎟ ⎨
−
+
−
⎝ 2⎠ ⎪ 1 ⎛ 1⎞
3 1 ⎛ 1 ⎞ ⎜⎝ 2 ⎟⎠
5 3 1 ⎛ 1 ⎞ ⎜⎝ 2 ⎟⎠
7 5 3 1 ⎛ 1⎞
2! ⋅ ⋅ Γ ⎜ ⎟
3! ⋅ ⋅ ⋅ Γ ⎜ ⎟
Γ ⎜ ⎟ 1! ⋅ Γ ⎜ ⎟
2 2 ⎝ 2⎠
2 2 2 ⎝ 2⎠
2 2 2 2 ⎝ 2⎠
⎩⎪ 2 ⎝ 2 ⎠
1
2
=
x
1
⋅
2 ⎛ 1⎞
Γ⎜ ⎟
⎝ 2⎠
⎧ 2 2x2 2x 4 2x6
⎫
+
−
+ ...⎬ =
⎨ −
1!
3!
5!
7!
⎩
⎭
⎫
⎪⎪
⎛ x⎞
⋅ ⎜ ⎟ + ...⎬
⎝ 2⎠
⎪
⎭⎪
6
⎫
2 1 ⎧ x x3 x5 x7
2
sin x.
⋅
⎨ − + − + ...⎬ =
x π ⎩1! 3! 5! 7!
x
π
⎭
14. (c) State Cayley–Hamilton theorem and so that matrix
⎡ 0 0 1⎤
A = ⎢⎢ 3 1 0 ⎥⎥ satisfies the above theorem :
⎢⎣ −2 1 4 ⎥⎦
Ans. Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation.
⎡ 0 0 1⎤
We have A = ⎢⎢ 3 1 0 ⎥⎥
⎣⎢ −2 1 4 ⎥⎦
Characteristic equation of A
−λ
0
3 1− λ
−2
1
1
0 =0
4−λ
Q.16

Engineering Mathematics-II
or,
−λ{(1 − λ ) (4 − λ ) − 0}0 + 1{3 + 2(1 − λ )} = 0
or,
λ 3 − 5λ 2 + 6λ − 5 = 0
Now,
⎡ 0 0 1⎤
⎡ 0 0 1 ⎤ ⎡ 0 0 1 ⎤ ⎡ −2 1 4 ⎤
⎢
⎥
2
A = ⎢ 3 1 0 ⎥ , A = A. A = ⎢⎢ 3 1 0 ⎥⎥ ⎢⎢ 3 1 0 ⎥⎥ = ⎢⎢ 3 1 3 ⎥⎥
⎢⎣ −2 1 4 ⎥⎦
⎢⎣ −2 1 4 ⎥⎦ ⎢⎣ −2 1 4 ⎥⎦ ⎢⎣ −5 5 14 ⎥⎦
⎡ −2 1 4 ⎤ ⎡ 0 0 1 ⎤ ⎡ −5 −5 14 ⎤
A3 = A2 ⋅ A = ⎢⎢ 3 1 3 ⎥⎥ ⎢⎢ 3 1 0 ⎥⎥ = ⎢⎢ −3 4 15⎥⎥
⎢⎣ −5 5 14 ⎥⎦ ⎢⎣ −2 1 4 ⎥⎦ ⎢⎣ −13 9 51⎥⎦
3
2
A − 5 A + 6 A − 51
⎛ −5 5 14⎞
⎛ −2 1 4 ⎞
⎛ 0 0 1⎞
⎛ 1 0 0⎞
= ⎜ −3 4 15⎟ − 5 ⎜ 3 1 3 ⎟ + 6 ⎜ 3 1 0⎟ − 5 ⎜ 0 1 0⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎜
⎟
⎝ −13 19 51⎠
⎝ −5 5 14⎠
⎝ −2 1 4⎠
⎝ 0 0 1⎠
⎛ 0 0 0⎞
= ⎜ 0 0 0⎟ = 0
⎜
⎟
⎝ 0 0 0⎠
∴ A satisfies its characteristic equation.
So, A satisfies the Cayley–Hamilton theorem.
ENGINEERING MATHEMATICS—YEAR 2008
GROUP–A
(Multiple Choice Type Questions)
1. Choose the correct alternatives for any ten of the following:
(i)
1
x 2 is equal to
D −1
(a) x 2 + 2 x + 2
(b) −( x 2 + 2 x + 2)
(c) 2x − x 2
(d) −(2 x − x 2 )
Ans. −( x 2 + 2 x + 2)
Hint:
1
( x 2 ) = −(1 − D ) −1 x 2 = −(1 + D + D 2 + D 3 + ...) x 2 = −( x 2 + 2 x + 2 + 0 + ...)
D −1
(ii) An integrating factor for the differential equation
(a) e t
(b)
(c) et
(d)
dy
+ y = 1 is
dt
e
t
t
e
Ans. (a) e t
Hint:
dy
+ y =1
dt
I .F . = e ∫ = e t
dt
(iii) The general solution of the differential equation D 2 y + 9 y = 0 is
(a) Ae 3 x + Be −3 x
(b) ( A + Bx )e 3 x
(c) A cos 3 x + B sin 3 x
(d) ( A + Bx ) sin 3 x
Ans. (c) A cos 3 x + B sin 3 x
Hint: ( D 2 + 9) y = 0.
∴ m2 + 9 = 0
m = ±3i
C.F. = e 0, x { A cos 3 x + B sin 3 x}
∴ y = A cos 3 x + B sin 3 x
Q.18

Engineering Mathematics-II
1
(iv) If L{ f (t )} = tan −1 ⎛⎜ ⎞⎟ , then L{tf (t )} is
⎝ P⎠
⎛ 1 ⎞
(a) tan −1 ⎜ 2 ⎟
⎝P ⎠
(c)
Ans. (b)
(b)
1
1+ P2
(d) tan −1 ⎛⎜ 2 ⎞⎟
⎝ πP⎠
1
1+ P
1
1+ P2
L{ f (t )} = tan −1
1
cot −1 p
p
Now L{tf (t )} = −
d
1
(cot −1 p) =
dp
1 + p2
(v) In Simpson’s 1/3 rule of finding
b
∫ f ( x)dx, f ( x)
a
(a) line segment
(c) circular sector
is approximated by
(b) parabola
(d) parts of ellipse
Ans. (b) parabola.
(vi) The system of equations x + 2 y = 5, 2 x + 4 y = 7 has
(a) unique solution (b) no solution
(c) infinite number of solutions
Ans. (b) no solution.
(d) none of these
x + 2y = 5
2x + 4 y = 7
The equations can be written as AX = B
1 2 : 5⎞
Augmented Matrix Ab = ⎛
⎜⎝ 2 4 : 7⎟⎠
⎛1 2 : 5 ⎞
⎜⎝ 0 0 : −3⎟⎠ R′ 2 → R2 → 2 R1
Rank of augmented matrix is 2 Rank of coefficient matrix is 1 which are not equal.
Hence equations have no solution.
(vii) The four vectors (1, 1, 0, 0), (1, 0, 0, 1), (1, 0, a, 0), (0, 1, a, b) are linearly independent if
(a) a ≠ 0, b ≠ 2
(b) a ≠ 2, b ≠ 0
(c) a ≠ 0, b ≠ −2
(d) a ≠ −2, b ≠ 0
Solved Question Papers
Ans. (c) a ≠ 0, b ≠ −2
1 1 0 0
1 0 0 1
1 0 a 0
0 1 a b
= − a − ( ab + a)
= −2a − ab
= − a(2 + b)
∴ | A |≠ 0 i.e. a ≠ 0 and b ≠ −2.
⎛ 1 2⎞
(viii) The rank of a matrix A = ⎜
is
⎝ 2 3⎟⎠
(a) 1
(c) 3
(b) 2
(d) none of these.
Ans. (b) 2.
det A =
1 2
= −1 ≠ 0
2 3
∴ Rank is 2.
⎛
⎜
(ix) The matrix A = ⎜
⎜
⎜⎝
1
2
1
−
2
1 ⎞
2⎟
⎟
1 ⎟
⎟
2 ⎠
(a) an orthogonal
(b) a symmetric matrix
(c) an idempotent matrix
(d) a null matrix
Ans. (a) an orthogonal
⎛ 1
⎜ 2
AT A = ⎜
⎜− 1
⎜⎝
2
1 ⎞⎛
2⎟ ⎜
⎟⎜
1 ⎟⎜
⎟⎜
2⎠ ⎝
1
2
1
−
1 ⎞
1 ⎛ 1 1⎞ 1 ⎛1 −1⎞
2⎟
⎟=
⎜
⎟
⎜
⎟
1 ⎟
2 ⎝ −1 1⎠ 2 ⎝1 1 ⎠
⎟
2 ⎠
2
1 ⎛ 2 0⎞ ⎛ 1 0⎞
= ⎜
=
= I2
2 ⎝ 0 2⎟⎠ ⎜⎝ 0 1⎟⎠

Q.19
Q.20

Engineering Mathematics-II
(x) T : R2 → R2 is defined by T ( x1 , x2 ) = (2 x1 − x2 , x1 + x2 ). Then kernel of T is
(b) {(1, −1)}
(d) {(1, 2)},(1, −1)
(a) {(1, 2)}
(c) {(0, 0)}
Ans. (c) {(0, 0)} .
We know T ( x1 , x2 ) = (2 x1 − x2 , x1 + x2 ),( x1 , x2 ) ∈ R2
Let, ( x1 , x2 ) ∈ ker T
Then, T ( x1 , x2 ) = (0, 0)
⇒ 2 x1 − x2 = 0
and x1 + x2 = 0
⇒ x1 = 0, x2 = 0
(xi) Δ 2 e x is equal to (where h = 1)
(a) (e − 1)2 e x
(c) e 2 x (e − 1)
(b) (e − 1)e x
(d) e 2 x +1
Ans. (a)
Δ 2 e x = Δ( Δ 2 e x )
{
}
= Δ e x +1 − Δe x {∵ Δ f ( x ) = f ( x + h) − f ( x )}
= Δe x +1 − Δe x , as h = 1
(
) (
(e − 2e + 1)
= e x + 2 − e x +1 − e x +1 − e x
= ex
)
2
= e x (e − 1)
2
⎛ 1 0 0⎞
(xii) The eigen value of the matrix ⎜ 0 2 0⎟ is
⎜
⎟
⎝ 1 0 3⎠
(b) 1, 2, 3
(a) 0, 0, 1
(c) 2, 3, 6
(d) none of these.
Hint: Let λ be the eigen value of the matrix A
∴ the characteristic equation is
or,
1− λ
0
0
0
2−λ
0 =0
1
0
3−λ
or, (1 − λ ) (2 − λ ) (3 − λ ) = 0
or,
λ = 1, 2,3
(xiii) The mapping f : R → R where f ( x ) = sin s, x ∈ R is
(a) one-one
(b) onto
(c) neither one-one nor onto
(d) both one-one and onto
Ans. (c) neither one-one nor onto.
Solved Question Papers
GROUP—B
2. Solve the differential equation by Laplace transformation.
d2 y
+ 9y = 1
dt 2
Taking Laplace transform on both side
1
s 2Y ( s) − sy(0) − y ′(0) + 9Y ( s) =
s
1
or, x 2 y( s) − s.1 − a + 9Y ( s) = [Let y ′(0) = a]
s
1
2
or, ( s + 9) Y ( s) = a + s +
s
a
s
1
∴ Y ( s) = 2
+ 2
+
2
s + 9 s + 9 s( s + 9)
d2 y
⎛π⎞
+ 9 y = 1 y(0) = 1, y ⎜ ⎟ = −1.
2
⎝ 2⎠
dt
Ans. We have
=
a
s
1⎛1
s ⎞
+
+ ⎜ −
⎟
s2 + 9 s2 + 9 9 ⎝ s s2 + 9 ⎠
∴ y(t ) = L−1{Y ( s)}
a
1
sin 3t + cos 3t + (1 − cos 3t )
3
9
1 a
8
= + sin 3t + cos 3t
9 3
9
π
1
a
⎛ ⎞
∴ y ⎜ ⎟ = + ( −1) + 0 = −1
⎝ 2⎠ 9 3
10
or, a =
3
∴ required solution is
1 10
8
y(t ) = + sin 3t + cos 3t
9 9
9
=
3. Solve by the method of variation of parameters
d2 y
+ 9 y = sec 3 x.
dx 2
Ans. Refer to 3. (a) of 2005.
4. Examine the consistancy of the following system of equations and solve if possible.
x+ y+ z =1
2x + y + 2z = 2
3x + 2 y + 3z = 5
Ans. Refer to Q. 3. (ii) a of 2006.
5. If W = {( x, y, z ) ∈ R3 ; x + y + z = 0}, show that W is a subspace of R3, and find a basis of W.
Ans. Same as Q. 3. (b) of 2005.

Q.21
Q.22

Engineering Mathematics-II
6. Examine whether the mappint T : R2 → R defined by T ( x, y ) = (2 x − y, x ) is linear.
Ans. Let α = (α1 , α 2 ) and β = (β1 , β2 ) be two (???) R2
T (α ) = T (α1 , α 2 ) = (2α1 − α 2 , α1 )
Now,
T (β ) = T (β1 , β2 ) = (2β1 − β2 , β1 )
T ( aα + bβ ) = T {a(α1 , α 2 ) + b(β1 , β2 )}
= T {aα1 + bβ1 , aα 2 + bβ2 }
= {2( aα1 + bβ1 ) − aα 2 − bβ2 , aα1 + bβ1}
= a(2α1 − α 2 , α1 ) + b(2β1 − β2 , β1 )
= aT (α ) + bT (β )
Now,
∴
Given T is a linear transformation.
1
7. Evaluate
dx
∫ 1+ x
using Trapezoidal rule, taking four equial sub-intervals.
0
1
; a = 0, b = 1, x = 4
1+ x
b − a 1− 0 1
h=
=
= = 0.25
n
4
4
Ans. We have f ( x ) =
So,
x
0
0.25
.5
.75
1
f (x)
1
0.8
0.6666
0.5714
0.5
from, Trapezoidal rule.
1
dx
h
= [ y0 + 2( y1 + y3 + y3 ) + y4 ]
1+ x 2
0
I =∫
0.25
[1 + 2(0.8 + 0.6666 + 0.5714) + 0.5]
2
= 0.695 Ans.
=
GROUP—C
8. (a) Find Laplace transform of f (t ) = sin t , 0 < t < π = 0, t > π
Ans. Same as Q. 8. (b) of 2004.
⎡1 0 2⎤
(b) It A = ⎢⎢0 1 1 ⎥⎥ then verify that A satisfies its own characteristic equation. Hence find A−1 .
⎢⎣0 1 0 ⎥⎦
Ans. Same as Q. 4. (a) of 2002.
Solved Question Papers
(c) Show that (3, 1, –2), (2, 1, 4), (1, –1, 2) form a basis of R3.
3 1 −2
Ans. As 2 1 4
1 −1 2
= 3{2 + 4} − 1{4 − 4} − 2{−2 − 1}
= 18 − 0 + 6
= 24 ≠ 0
Given three vectors are linearly independent so the vectors belong to R3.
∴ The vectors form a basis.
d2 y
dy
− x − 3 y = x 2 log x
2
dx
dx
Ans. Let, log x = z
9. (a) Solve: x 2
dy 1 dy
= ⋅
dx x dz
d2 y
1 dy 1 d 2 y dz
∴
=
−
⋅ +
dx 2
x dz x dz 2 dx
1 dy 1 d 2 y
=−
+ ⋅
x dz x 2 dz 2
dy dy
Naturally x
=
dx dz
d 2 y d 2 y dy
x2 2 = 2 −
dx
dz
dz
Putting these values in given equation.
∴
d2 y
dy
− 2 ⋅ − 3 y = ze 2 z
2
dz
dz
The auxileary equation is
m2 − 2 m − 3 = 0
or, ( m − 3) ( m + 1) = 0
∴ C.F. = A11e 3 z + A22 e − z , A11 and A22 constants
1
P.I. = 2
( ze 2 z )
D − 2D − 3
1
( z)
= e2 z ⋅
( D + 2)2 − 2( D + 2) − 3
1
( z)
= e2 z ⋅ 2
D + 4 D + 4 − 2D − 4 − 3
1
( z)
= e2 z ⋅ 2
D + 2D − 3
⎛ D2 + 2D ⎞
1
= − ⋅ e 2 z ⎜1 −
⎟⎠
3
3
⎝
−1
z

Q.23
Q.24

Engineering Mathematics-II
⎧⎪ D 2 + 2 D ⎛ D 2 + 2 D ⎞ 2
1
⎪⎫
= − ⋅ e 2 z ⎨1 +
+⎜
+ ...⎬ z
⎟
3
3
3
⎝
⎠
⎩⎪
⎭⎪
1
1
= − ⋅ 3e 2t (3 z + 2) = − e 2t (3 z + 2)
3
9
∴ Total solution is
1
= A11e 3 z + A22 e − z − e 2 z (3 z + 2)
9
1
= A11 x 3 + A22 . x −1 − x 2 (3log x + 2) [ as z = log x ]
9
d
9. (b) Solve: ( D 2 + 4) y = x sin 2 x, which D =
dx
Ans. Auxiliary equation 0 is
m2 + 4 = 0
∴ m2 = −4 = (2i )2
∴ m = ± 2i
∴ C.F. = A1 cos 2 x + A2 sin 2 x
1
x sin 2 x
P.I. = 2
D +4
1 1
x(1 − cos 2 x )
=
2 D2 + 4
1 1
1 1
( x) −
=
⋅ x cos x
2 D2 + 4
2 D2 + 4
1
1 ⎛
1
⎞ 1
cos 2 x
= ( D 2 + 4) −1 ( x ) − ⋅ ⎜ x − 2
⋅ 2 D⎟ 2
⎠ D +4
D +4
2
2 ⎝
1
=
1 ⎛ D2 ⎞
1⎛
2D ⎞ 1
1+
( x) − ⎜ x − 2
cos 2 x
⎟x
D + 4 ⎠ 2D
2.4 ⎜⎝
4 ⎟⎠
2⎝
=
x 1⎛
2D ⎞ 1
− ⎜x− 2
⎟ x sin 2 x
D + 4⎠ 2
8 4⎝
=
x 1⎛
2D ⎞
− ⎜x− 2
⎟ x sin 2 x
D + 4⎠
8 8⎝
x x 2 sin 2 x 1 1
1 1 1
−
−
sin 2 x − −
( x cos 2 x )
2
8
8
4 D +4
2 2 D2 + 4
1
If I = 2
( x cos 2 x )
D +4
x 2 sin 2 x 1
1
=
+ ⋅ x ⋅ sin 2 x − 1
4
2
2
x 2 sin 2 x x cos 2 x
∴ 2I =
+
4
8
1 2
x cos 2 x
∴ I = x sin 2 x +
8
16
1
1
=
Solved Question Papers
1 2
1
x sin 2 x +
x cos 2 x
16
32
∴ general solution : C.F. + P.I.
1
1
= A1 cos 2 x + A2 sin 2 x − x 2 sin 2 x − x cos 2 x.
16
32
dy
9. (c) Solve, x + y = y 2 log x
dx
dy
Ans. x + y = y 2 log x
dx
∴ P.I. =
or,
x dy 1
+ = log x
y 2 dx y
1
=z
y
Let,
1 dy dz
=
y 2 dx dx
dz
∴ − x + z = log x
dx
dz z
log x
or,
− =−
..........(i)
dx x
x
∴−
As it is linear equation of 2
∴ I.F. = e
−
1
∫ x dx
= e − log| x| =
1
x
Multiplying both sides of (i) by I.F. =
1
x
log x
1 dz 1
. − 2 .z = 2
x dx x
x
or,
∴
log x
d ⎛ z⎞
⎜⎝ ⎟⎠ = − 2
dx x
x
log x
z
= − ∫ 2 dx
x
x
= − ∫ pe − p dp
Let,
= ( P + 1)e − P + c. c = constant
∴
1
(log x + 1) + c
x
[as P = log x ]
∴ x = eP
=
log x = P
1
dx = dp
x
10. (a) Compute f (0.5) and f (0.9) from the following table
x
0
1
2
3
f(x)
1
2
11
34

Q.25
Q.26

Engineering Mathematics-II
Ans. Required difference table is
f ( x) Δ f ( x) Δ 2 f ( x) Δ3 f ( x)
1
1
2
8
9
6
11
14
23
34
x
0
1
2
3
to evaluate f (0 ⋅ 5) using Newton’s Farward interpolation formula where
x − x0 0 ⋅ 5 − 0
x = 0 ⋅ 5, x0 = 0, h = 1, d =
=
= 0⋅5
h
1
∴ f ( x ) = f ( x0 ) + d Δf ( x0 ) +
= 1 + 0 ⋅ 5.1 +
= 0 ⋅ 875
d ( d − 1)
2!
0 ⋅ 5 (0 ⋅ 5 − 1)
2
.8 +
Δ 2 f ( x0 ) +
d ( d − 1)( d − 2)
3!
0 ⋅ 5 (0 ⋅ 5 − 1)(0 ⋅ 5 − 2)
6
Δ 2 f ( x0 )
.6
Again for f (0 ⋅ 9) , x0 = 0 ⋅ 9, x0 = 0, h = 1
d=
∴ f (0 ⋅ 9) = 1 + 0 ⋅ 9.1 +
x − x0 0 ⋅ 9 − 0
=
= 0⋅9
h
1
0 ⋅ 9 (0 ⋅ 9 − 1)
2!
.8 +
⎧
1
(b) Apply convolution theorem to evaluate L−1 ⎪⎨
2
⎪⎩ s + 2 s + 5
1
1
⎧
⎫
Ans. We know L−1 ⎨ 2
= sin 2t
2 ⎬
⎩s + 2 ⎭ 2
(
)
2
0 ⋅ 9 (0 ⋅ 9 − 1)(0 ⋅ 9 − 2)
6
⎫
⎪
⎬
⎪⎭
⎧⎪
⎫⎪ 1 −1
1
∴ L−1 ⎨
⎬ = e sin 2t
2
2
⎪⎩ ( s + 1) + 2 ⎪⎭ 2
1
⎧
⎫ 1 −1
∴ L−1 ⎨ 2
⎬ = e sin 2t
⎩ s + 2s + 5 ⎭ 2
∴ Using convoluation theorem
⎧
1
⎪
L−1 ⎨
2
⎩⎪ s + 2 s + 5
(
⎫ 1
2 (t − u )
⎪
− u sin 2u − (t − u ) sin
=
e
e
du
⎬
∫
2
2
2
⎭⎪ 0
)
.6
Solved Question Papers
e −t
8
∫ ⎡⎣cos (4u − 2t ) − cos 2t ⎤⎦ du
e −t
=
8
⎡ sin ( 4u − 2t )
⎤
− u cos 2t ⎥
∫0 ⎢⎣
4
⎦0
=
=
10. (c) Prove that
1
0
1
t
e − t ⎛ sin 2t
⎞
− t cos 2t ⎟
⎠
8 ⎜⎝ 2
( x − a)2 ( x − b)2 ( x − c )2
( y − a)2 ( y − b)2 ( y − c )2
( z − a)2 ( z − b)2 ( z − c )2
= 2 ( x − y )( y − z )( z − x )( a − b )(b − c )(c − a )
Ans. Let,
Δ=
( x − a)2 ( x − b)2 ( x − c )2
( y − a)2 ( y − b)2 ( y − c )2
( z − a)2 ( z − b)2 ( z − c )2
x2
= y2
z2
−2 x 1 1
−2 y 1 a
−2 z 1 a 2
1
b
b2
1
x2
c = −2 y 2
c2
z2
x 1 1
y 1 a
z 1 a2
1
b
b2
1
c
c2
= −2 Δ1 Δ 2
Now,
x2
Δ1 = y 2
z2
x 1
x2
y 1 = y2 − x2
z2 1 z2 − x2
x2
= ( y − x )( z − x ) y 2 − x 2
z2 − x2
x
1 1
⎡ R = R2 − R1 ⎤
y−x 0 ⎢ 2
⎥
R3 = R2 − R1 ⎦
⎣
z−x 0
x
1
y−x 0
z−x 0
= ( y − x )( z − x ){ y − x − z + x} = − ( x − y )( y − z )( z − x )
1
Similarly Δ 2 = a
a2
1
b
b2
1
c = ( a − b )(b − c )(c − a )
c2
∴Δ = Δ1 Δ 2 = 2 ( x − y )( y − z )( z − x )( a − b )(b − c )(c − a ) (Proved)
11. (a) Prove that Δ. ∇ = Δ − ∇, where symbols have their usual meaning.
∇ ⎡⎣ f ( x )⎤⎦ = f ( x ) − f ( x − h)

Q.27
Q.28

Engineering Mathematics-II
}
∴Δ. ∇ ⎣⎡ f ( x )⎦⎤ = Δf ( x ) − Δf ( x − h) = Δf ( x ) − ⎡⎣ f ( x ) − f ( x − h)
= Δf ( x ) − ∇f ( x ) = ( Δ − ∇) f ( x )
∴Δ. ∇ = Δ − ∇
(b) Estimate missing term in the following table:
x
0 1 2 3 4
f ( x ) 1 3 9 − 31
4
Ans. Here y has four given values. Naturally Δ y0 = 0
or,
or,
( E − 1)4
(E
4
y0 = 0
)
− 4 E 3 + 6 E 2 − 4 E + 1 y0 = 0
y4 − 4 y3 + 6 y2 − 4 y1 + y0 = 0
y + 6 y2 − 4 y1 + y0 81 + 6.9 − 4.3 + 1
∴ y3 = 4
=
4
4
= 31
or,
(
)
(
)
)
(
11. (c) Solve: x 2 y − 2 xy 2 dx + 3 x 2 y − x 3 dy = 0
(
)
Ans. We have x 2 y − 2ky 2 dx + 3 x 2 y − x 3 dy = 0........(i)
∴ M = x y − 2 xy
2
2
N = 3x y − x
2
∴
∂M
∂N
= x 2 − 4 xy
= 6 xy − 3 x 2
∂y
∂x
∴
∂M ∂N
−
= − 10 xy − 4 x 2 ≠ 0
∂y
∂x
(
3
)
∴ It is not a exact equation.
Now Mx + Ny = x 2 y − 2 xy 2 x + 3 x 2 y − x 3 y = x 2 y 2
(
So the integrating factor
) (
1
x2 y2
Multiplying both sides of (i) by
1
x2 y2
x 2 y − 2 xy
3 x 2 y − yx 3
dx +
dy = 0
2 2
x y
x2 y2
or,
ydx − xdy 2
3
− dx + dy = 0
2
x
y
y
)
Solved Question Papers
Inlogration both sides we get
or,
x
− 2 log x + 3 log y = c, c → constant
y
y3
x
+ log 3 = c Ans.
y
x
12. (a) Show that
{
}
d
x n J n ( x ) = x n J n −1 ( x )
dx
Ans. We know,
⎛ x⎞
J n ( x ) = ∑ ( −1) ⎜ ⎟
⎝ 2⎠
r=0
∞
∞
∴ x n J n ( x ) = ∑ ( −1)
r=0
r
2
x + 2r
n+ 2r
r
1
r ! Γ ( n + r + 1)
x 2 x + 2r
r ! Γ ( x + r + 1)
Differentiating both side w.r.t. x
α
d
2 x + 2r. x 2 x + 2 r −1
r
⎡⎣ x n J n ( x )⎤⎦ = ∑ ( −1) . x + 2 r
dx
2
r ! Γ ( n + r + 1)
r=0
n + 2 r −1
α
1. ( x + r )
r ⎛ x⎞
= ∑ ( −1) . ⎜ ⎟
.x n
⎝
⎠
2
r
!
Γ
x
+
r
!
)
(
r=0
(n + r )
α
r ⎛ x⎞
= x n ∑ ( −1) . ⎜ ⎟
⎝ 2⎠
r=0
n + 2 r −1
1
Γ (x + r)
= xn J n −1 ( x ) (Proved)
12. (b) Obtain a function whose 1st difference is 9 x 2 + 11x + 5, where h = 1
Ans. Let the required function is f ( x )
∴ Δf ( x ) = 9 x 2 + 11x + 5
1
⎡9 x 2 + 11x + 5⎤⎦
Δ⎣
1
= ⎡⎣9( x (2) + x (1) + 11x (1) + 5⎤⎦
Δ
1
∴ f ( x ) = ⎡⎣9 x (2) + 20 x (1) + 5⎤⎦
Δ
f ( x) =
x (3)
x (2)
+ 20.
+ 5 x (1) + k
[ k = constant ]
3
2
⎛ 9⎞
⎛ 20 ⎞
= ⎜ ⎟ x ( x − 1)( x − 2) + ⎜ ⎟ x ( x − 1) + 5 x + k
⎝ 3⎠
⎝ 2⎠
= 9.
= 3 x ( x − 1)( x − 2) + 10 x ( x − 1) + 5 x + k
= 3 x ( x − 1)( x − 2) + 5 x {2 x − 2 + 1} + k
= 3 x ( x − 1)( x − 2) + 5 x (2 x − 1) + k
= 3x 3 + x 2 + x + k

Q.29
Q.30

Engineering Mathematics-II
12. (e) Solve the differential equation by Laplace transform—
d 2 y 2dy
+
+ 5 y = e t sin t ; y(0) = 0 y ′(0) = 1
dx
dx 2
Ans. We have
d2 y
dy
+ 2. + 5 y = e t sin t
dx
dx 2
Taking Laplace transform on both side
s 2Y ( s ) − sy (0 ) − y ′ (0 ) + 2 ⎡⎣ sY ( s ) − y (0 )⎤⎦ + 5Y ( s ) =
1
( s − 1)2 + 1
Putting y (0 ) = 0 and y′ (0 ) = 1
s 2Y ( s ) − 0 − 1 + 2 sY ( s ) − 0 + 5Y ( s ) = 1
s 2Y ( s ) − 0 − 1 + 2 sY ( s ) − 0 + 5Y ( s ) =
or,
(s
2
)
+ 2s + 5 Y ( s) = 1 +
1
( s − 1)2 + 1
1
( s − 1)2 + 1
Y ( s) =
1
1
+
2
s + 2 s + 5 [( s − 1) + 1][ s 2 + 2 x + 5]
=
1
1
+
2
2
( s + 1) + 4 [( s − 1) + 1][( s + 1)2 + 4]
2
⎤
4 ( s − 1)
1 ⎡ 4 ( s + 1)
66
7
+
−
+
⎢
⎥
2
2
2
2
65 ⎢⎣ ( s + 1) + 22 ( s + 1) + 22 ( s − 1) + 1 ( s − 1) + 1⎥⎦
∴ y ( f ) = L−1 {Y ( s )}
=
{
}
1
4e − t cos 2t + 33e − t sin 2t − 4e t cos t + 7e t sin t .
65
13. (a) What is meant by linear independence of a set of n-vectors?
=
Show that the vectors (1, 2,1) , ( −1,1, 0 ) and (5, −1, 2) linearly independent.
Ans. Assume V be a vector over a field F.
A finite subset { x1 , x2 , ..., xn } of vectors of v is said to be linearly independent if every relation satisfies
a1 x1 + a2 x2 + ... + an xn = 0, ai ∈ F
1 ≤ i ≤ x ⇒ ai − 0 for each 1 ≤ i ≤ n.
1 2 1
−1 1 10 = 1 {2 − 0} − 2 {−2 − 0} + 1 {1 − 5} = 2 ≠ 0
5 −1 2
∴ Vectors are linearly independent.
Solved Question Papers
{
}

Q.31
(b) Show that W = ( x, y, z ) ∈ R3 : 2 x − y + 3 z = 0 is a subspace of R 3. Find a basis W. What is its dimension?
Ans. Same as Q. 3, (a) of 2002.
(b + c )2
a2
a2
b2
(c + a)2
b2
c2
c2
(a + b )2
Δ2 =
(c) Prove that
= 2abc ( a + b + c )
3
Ans. Same as Q.2(a) of 2004.
14. (a) Find the rank of the matrix.
−1
2
0
1
Ans. Same as Q.2. (b) of 2002.
14. (b) Solve by Cramer’s Rule
2 −1
4 4
0 1
6 3
2x − y = 3
3y − 2x = 5
−2 z + x = 4
Ans. Here,
⎡ 2 −1 0 ⎤
Δ = ⎢⎢0 3 −2⎥⎥ = −10
⎢⎣1 0 −2⎥⎦
⎡ 3 −1 0 ⎤
Δ1 = ⎢⎢ 5 3 −2⎥⎥ = −20
⎢⎣ 4 0 −2⎥⎦
⎡2 3 0 ⎤
Δ 2 = ⎢⎢0 5 −2⎥⎥ = −10
⎢⎣1 4 −2⎥⎦
⎡ 2 −1 3 ⎤
Δ 3 = ⎢⎢0 3 5 ⎥⎥ = 10
⎢⎣1 0 4 ⎥⎦
Δ1 −20
=
=2
Δ −10
Δ
−10
y= 2 =
=1
Δ
−10
Δ
10
z= 3 =
= −1
Δ
−10
∴x =
0
2
5
2
Q.32

Engineering Mathematics-II
x = 2⎫
∴ Required solution is y = 1⎪⎬ Ans.
z = 1 ⎪⎭
⎡1 −1 2 ⎤
⎢
⎥
14. (c) Find the eigen values and corresponding eigen vectors of the matrix A = ⎢2 −2 4 ⎥
⎢⎣ 3 −3 6 ⎥⎦
Ans. Characteristic equation of given matrix is
1− λ
2
3
or,
or,
−1
2
−2 − λ
4 =0
−3
6−λ
(1 − λ ) {(6 − λ ) − ( −2 − λ )} + 1. {2 (6 − λ ) − 4.3} + 2 {−6 + (2 + λ ) .3} = 0
λ 2 (5 − λ ) = 0
∴ λ = 0, 0, 5
for λ = 0 let there be a non-zero eigen vector X.
⎛ 1 −1 2⎞ ⎛ x1 ⎞ ⎛ 0⎞
∴ ⎜ 2 −2 4⎟ ⎜ x2 ⎟ = ⎜ 0⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
⎝ 3 −3 6⎠ ⎝ x3 ⎠ ⎝ 0⎠
⎛ 1 −1 2⎞ ⎛ x1 ⎞ ⎛ 0⎞
R21 → R2 − 2 R1 ⎜
⎯⎯⎯⎯⎯⎯⎯
→
0 0 0 ⎟ ⎜ x2 ⎟ = ⎜ 0 ⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
R31→ R3 −3R1
⎝ 0 0 0⎠ ⎝ x3 ⎠ ⎝ 0⎠
∴ x1 − x2 + 2 x3 = 0
⎛ 1⎞
⎛ −2⎞
⎜
⎟
∴ Required region vectors for ( λ = 0 ) are A 1 and B ⎜ 0 ⎟
⎜ ⎟
⎜ ⎟
⎝ 0⎠
⎝ 1⎠
A and B non-zero real number.
⎛ −4 −1 2⎞ ⎛ x1 ⎞ ⎛ 0⎞
Again for λ = 5 ⎜ 2 −7 4⎟ ⎜ x2 ⎟ = ⎜ 0⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
⎝ 3 −3 1⎠ ⎝ x3 ⎠ ⎝ 0⎠
⎛ −1 −4 3⎞ ⎛ x1 ⎞ ⎛ 0⎞
R11 = R1 + R3 ⎜
⎯⎯⎯⎯⎯⎯→ 2 −7 4⎟ ⎜ x2 ⎟ = ⎜ 0⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
⎝ 3 −3 1⎠ ⎝ x3 ⎠ ⎝ 0⎠
R21 → R2 − 2 R1 ⎛ −1 −4 3 ⎞ ⎛ x1 ⎞ ⎛ 0⎞
⎯⎯⎯⎯⎯⎯⎯
→ ⎜ 0 −15 10⎟ ⎜ x2 ⎟ = ⎜ 0⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
1
R3 → R3 −3R1
⎝ 0 −15 10⎠ ⎝ x3 ⎠ ⎝ 0⎠
⎛ −1 −4 3 ⎞ ⎛ x1 ⎞ ⎛ 0⎞
R31 = R3 + R2 ⎜
⎯⎯⎯⎯⎯⎯→
0 −15 10⎟ ⎜ x2 ⎟ = ⎜ 0⎟
⎜
⎟⎜ ⎟ ⎜ ⎟
0
1 ⎠ ⎝ x3 ⎠ ⎝ 0⎠
⎝3
Solved Question Papers
∴ − x1 − 4 x2 + 3 x3 = 0
−15 x2 + 10 x3 = 0
x2 x3
=
= C , C → constant non-zero real no.
2
3
Hence required legen veecler is
From these two equaliers
⎛ x1 ⎞
⎛ 1⎞
⎜ x ⎟ = C ⎜ 2⎟ .
⎜ 2⎟
⎜ ⎟
⎝ 3⎠
⎝ x3 ⎠

Q.33
ENGINEERING & MANAGEMENT EXAM—YEAR JUNE, 2009
SEMESTER—2
GROUP—A
Multiple Choice Type Questions
1. Choose the correct alternatives for any ten of the following:
⎡ 1 0⎤
100
(i) If A = ⎢
⎥ , then A is
⎣ −1 1 ⎦
⎡ 1 0⎤
0⎤
⎡ 1
(a) ⎢
(b) ⎢
⎥
⎥
⎣ −50 1 ⎦
⎣ −150 1 ⎦
0⎤
⎡ 1
(c) ⎢
⎥
⎣ −100 1 ⎦
(d) None of these
1− λ
0
⎛ 1 0⎞
2
Ans. A = ⎜
; A− λI = 0 ⇒
= 0 ⇒ (1 − λ ) = 0
⎟
1
1
1
1
λ
−
−
−
⎝
⎠
⇒ λ 2 − 2λ + 1 = 0
⇒ A2 − 2 A + I 2 = 0
∴ A2 = 2 A − I 2
∴ A4 = A2 . A2 = ( 2 A − I 2 )( 2 A − I 2 ) = 4 A2 − 4 A + I 2
= 8 A − 4 I2 − 4 A + I 2 = 4 A − 3I 2
A 3 = A2 ( A) = ( 2 A − I 2 ) A
= 2A2 − A
= 4 A − 2I 2 − A = 3 A − 2I 2
0 ⎞ ⎛ 99 0 ⎞ ⎛ 1
0⎞
⎛ 100
∴ A100 = 100 A − 99 I 2 = ⎜
−⎜
=⎜
⎟
⎟
⎝ −100 100⎠ ⎝ 0 99⎠ ⎝ −100 1⎟⎠
∴
Ans. (c)
(ii) The set of vectors {(2, 1, 1), (1, 2, 2), (1, 1, 1)} in R3 is—
(a) linearly dependent
(b) linearly independent
(c) basis of R3
(d) none of these
⎡2 1 1⎤
⎡1 1 1 ⎤ R2 − R1 ⎡1 1 1 ⎤
⎡1 1 1 ⎤
~
⎢1 2 2⎥
⎢0 1 1 ⎥ ~ ⎢0 1 1 ⎥
Ans. ⎢1 2 2⎥ R1 ←⎯
→
R
~
3 ⎢
⎢
⎥
⎥
⎢
⎥
⎢
⎥
⎢⎣1 1 1 ⎥⎦
⎢⎣2 1 1 ⎥⎦ R3 − 2 R1 ⎢⎣0 −1 −1⎥⎦ R3 + R1 ⎢⎣0 0 0 ⎥⎦
— Raw echelon form
rank = no of non zero rows
= 2 < no of unknowns = 3
∴ the set of vectors are linearly dependent.
Ans. (a)
10 × 1 = 10
Solved Question Papers
1 ⎤
⎡ 1
⎢ 2 − 2⎥
⎥ is
(iii) The matrix A = ⎢
1 ⎥
⎢ 1
⎢
⎥
2 ⎦
⎣ 2
(a) an orthogonal matrix
(c) an idempotent matrix
⎡
⎢
Ans. A = ⎢
⎢
⎢
⎣
1
1 ⎤
⎡ 1
⎥
2 ⎥ T ⎢⎢ 2
; A =
1 ⎥
⎢ 1
⎥
⎢−
2 ⎦
2
⎣
1
1 ⎤⎡ 1
−
2
2 ⎥⎥ ⎢⎢ 2
1
1 ⎥⎢ 1
⎥ ⎢−
2
2 ⎦⎣
2
1 ⎤
2 ⎥⎥
1 ⎥
⎥
2⎦
−
2
1
2
⎡
⎢
∴ A.T T = ⎢
⎢
⎢
⎣
(b) a symmetric matrix
(d) a null matrix
1 ⎤
2 ⎥⎥ ⎡1 0 ⎤
=
= I2
1 ⎥ ⎢⎣0 1 ⎥⎦
⎥
2⎦
∴ A is orthogonal matrix
Ans. (a)
1
(iv) The value of the determinant 12
0
(a) 0
(b) 1
4
22
1
16
4 2 is
6
(c) 4
(d) 22
4
22
1
16
4 2 = 0 as R1 and R2 are identical
6
1
Ans. 12
0
Ans. (a)
(v) The solution of a system of n linear equations with n unknowns is unique if and only if
(a) det A = 0
(b) det A > 0
(c) det A < 0
(d) det A ≠ 0.
where A is the matrix of the coefficients of the unknowns in the linear equations.
Ans. unique solution iff det A ≠ 0
∴ Ans. (d).
⎡1 4⎤
(vi) The eigen values of the matrix ⎢
⎥ are
⎣4 1⎦
(a) –5, –3
(b) –5, 3
(c) 3, –5
Ans.
A− λI = 0 ⇒
(d) 5, 3
1− λ
4
2
= 0 ⇒ (1 − λ ) − 16 = 0 ⇒ λ 2 − 2λ − 15 = 0 ⇒ λ = 5, − 3
4
1− λ

Q.35
Q.36

Engineering Mathematics-II
(vii) The general solution of p = log (px – y) where p =
(a) y = cx − c
(b) y = cx − e c
(c) y = c 2 x − e − c
(d) none of these
dy
is
dx
Ans. (b)
p = log ( px − y ) ⇒ px − y = e p ⇒ y = px − e p
(
)
dp
dp
dp
− ep
⇒ x − ep
= 0 ⇒ p = c or x = e p
dx
dx
dx
∴ general solution y = cx − e c
⇒ p = p.1 + x
(viii) Which of the following is not true (the notations have their usual meanings)?
(a) Δ = E − 1
Δ
(c)
= Δ+∇
∇
Ans. (c)
(b) Δ. ∇ = Δ − ∇
(d) ∇ = 1 − E −1
⎛ E − 1⎞
Δ + ∇ = ( E − 1) + 1 − E −1 = ( E − 1) + ⎜
⎝ E ⎟⎠
(
)
2
1⎞
⎛
⎛ E + 1⎞ ⎛ E − 1⎞
= ( E − 1) ⎜1 + ⎟ = ( E − 1) ⎜
=
⎝ E⎠
⎝ E ⎟⎠ ⎜⎝ E ⎟⎠
(ix) Δ 2 e x is equal to ( h = 1)
(a)
(c)
(e − 1)2 e x
e 2 x (e − 1)
(e − 1) e x
(b)
(d) e 2 x
Ans. (a)
( ( )) = Δ ( e
Δ2e x = Δ Δ e x
(
= e
= ex
x +1
2
(c)
π
3
4
x +1
∫
0
x+2
x
2
e
∞
π
) ( )
− e ) = e − 2e
− e x = Δ e x +1 − Δe x
) − e ) − (e
(e − 2 + 1) = (e − 1)
x+2
(x) The value of
(a)
x +1
x +1
+ ex
ex
sin t
dt is equal to
t
(b)
π
(d)
π
6
2
Ans. (d)
∞
sin t
⎛ sin t ⎞
⎛ 1⎞
⎛ 1⎞
Since L ⎜
= tan −1 ⎜ ⎟ ⇒ ∫ e − st
dt = tan −1 ⎜ ⎟ ;
⎟
⎝ t ⎠
⎝ s⎠
⎝ s⎠
t
0
Solved Question Papers
∞
Putting s = 0 we have
∫
0

Q.37
sin t
π
dt = tan −1 (∞ ) =
t
2
(xi) If S and T are two subspaces of a vector space V, then which one of the following is a subspace of
V also?
(a) S ∪ T
(b) S ∩ T
(c) S − T
(d) T − S
Ans. Since S ∩ T = {v : v ∈ s and v ∈T } and S, T are subspace of V
v1 , v2 ∈ S ∩ T ⇒ v1 , v2 ∈ S and v1 , v2 ∈T
⇒ C1V1 + C2V2 ∈ S and C1V1 + C2V2 ∈T
⇒ C1V1 + C2V2 ∈ S ∩ T
⇒ S ∩ T is a subspace of V
Ans. (b).
(xii) If λ 3 − 6λ 2 + 9λ − 4 is the characteristic equation of a square matrix A, then A-1 is equal to
1 2 3
9
(b)
(a) A2 − 6 A + 9 I
A − A+ I
4
2
4
1 2 3
9
2
(d) A − 6 A + 9
(c) A − A +
4
2
4
Ans. Every square matrix satisfies its own characteristic equation
⇒ A3 − 6 A2 + 9 A − 4 I = 0
(
)
⇒ A A2 − 6 A + 9 I = 4 I
(
1 2
A − 6 A + 9I
4
1
3
9
= A2 − A + I
4
2
4
⇒ A−1 =
)
−2 −3 4
(xiii) Co-factor of –3 in the determinant 1 0 1 is
0 −1 4
(a) 4
(b) –4
(c) 0
(d) none of these
Ans. (b)
Ans. Co-factor of ( −3) = ( −1)
1+ 2
Ans. (b)
1 1
= −4
0 4
Q.38

Engineering Mathematics-II
GROUP—B
(Multiple Choice Type Questions)
Answer any three of the following
3 × 5 = 15
2. If A be a skew symmetric and ( I + A) be a non-singular matrix, then show that B = ( I − A)( I + A) is
orthogonal.
−1
Ans. Given B = ( I – A)( I + A)
∴ BT B =
{( I + A) }
−1 T
( I − A)T ( I − A)( I + A)−1
= {( I + A) }−1 ( I − A)
T
(
= ( I + A)
(
= I + AT
= ( I − A)
T
T
)
−1
−1
( I − A)( I + A)−1
( I − A)T ( I − A)( I + A)−1
) ( I − A)( I + A)( I + A)
−1
−1
−1
[∵ AT = − A]
( I − A)( I + A)( I + A)−1
= I .I
= I (∵ A−1 A = 1)
∴ BT B = I ⇒ B is orthogonal matrix.
⎧⎪
⎫⎪
1
3. Evaluate: L−1 ⎨
2
3 ⎬
⎪⎩ ( s − 1) ( s − 2) ⎪⎭
1
A
B
C
D
E
=
+
+
+
+
Ans. Let
3
2
2
2
S
−
1
S
−
2
)
(
(S − 2) (S − 1)
(S − 1)
(S − 2) (S − 2)3
Where A, B, C, D, E are constants.
∴1 = A ( S − 1)( S − 1) + B ( S − 2) + C ( S − 1) ( S − 2) + D ( S − 2)( S − 1) + E (S − 1)
3
3
2
2
Let, S = 1 ⇒ B ( −1) ⇒ B = −1
Let, S = 2 ⇒ 1 = E ⇒ E = 1
equating coeff. of S3 from both sides.
3
0 = A + B + ( −6C ) + D
⇒ A − 6C + D = 1.....(i)
equation coeff. of S 4 and constant term from both sides
0 = A + C ⇒ A = −C
from (1) − 7C + D = 1 ...(ii)
1 = 8 A − 8 B + 4C − 2 D + E
or,
−8 = −4C − 2 D ⇒ 2C + D = 4 ....(iii)
2
2
Solved Question Papers
(ii) − (iv) − 9C = −3 ⇒ C =
∴ A= −
∴
1
3
1
2 10
∴D = 4− =
3
3
3
1
(S − 1) (S − 2)
2
3
= (−)
1
1
1
1
10
1
1
+ ( −1)
+ ⋅
+
+
2
2
3 ( S − 1)
(S − 1) 3 (S − 2) 3 (S − 2) (S − 2)3
⎡
⎤
1
1
⎛ 1 ⎞ 1 ⎛ 1 ⎞ 1 −1 ⎛ 1 ⎞
∴ L−1 ⎢
= − L−1 ⎜
+ L ⎜
⎟ −L ⎜
⎟
2
3⎥
2⎟
3 ⎝ S − 1⎠
⎝ ( S − 1) ⎠ 3 ⎝ S − 2 ⎠
⎣⎢ ( S − 1) ( S − 2) ⎦⎥
+
10 −1 ⎡ 1 ⎤ −1 ⎡ 1 ⎤
+L ⎢
L ⎢
2⎥
3⎥
3
⎣⎢ ( S − 2) ⎦⎥
⎣⎢ ( S − 2) ⎦⎥
1
1
10
t2
= − e t − e t t + e 2t + e 2t . t + e 2t .
3
3
3
2
4. Solve the differential equation—
dy
+ y = y 3 (cos x − sin x )
dx
dy
+ y = y 3 (cos x − sin x ) 1
dx
1 dy 1
⇒ 3
+
= (cos x − sin x )
y dx y 2
Ans.
1
2 dy dz
1 dy 1 dz
=z⇒− 3
=
⇒ 3
=
2
y
y dx dx
y dx 2 dx
1 dz
∴−
+ z = (cos x − sin x )
2 dx
dz
⇒
− 2 z = −2 (cos x − sin x ) ... ( I )
dx
− 2 dx
∴ I.F. = e ∫ = e −2 x
Let,
∴( I ) becomes on multiplication by I.F.
d
ze −2 x = −2e −2 x (cos x − sin x )
dx
⇒ d Ze −2 x = −2∫ e −2 x (cos x − sin x ) dx
(
(
)
)
⇒ Ze −2 x = −2 ⎡⎣ ∫ e −2 x cos dx − ∫ e −2 x sin xdx ⎤⎦
⎡ e −2 x
⎤
e −2 x
= −2 ⎢
( −2 cos x + sin x ) −
( −2sin x − cos x )⎥ + c
5
5
⎣
⎦
−2 x
= 2e
[3sin x − cos x ] + c
5

Q.39
Q.40

Engineering Mathematics-II
2
[3sin x − cos x ] + ce 2 x
5
1
2
∴ 2 = − (3sin x − cos x ) + ce 2 x
5
y
∴Z = −
where C is an arbitrary const.
4
5. Evaluate the definite integral
the error.
3
by using Trapezoidal rule, taking five ordinates and calculate
1
∴ f ( x ) = x + x3
a = 1, b = 4, nh = (b − a ) = 3
4
∫ ( x + x ) dx
3
Ans.
∫ ( x + x ) dx
1
gives there are 5 ordinates i.e., there are 4 subintgervals ⇒ n = 4
∴h =
3
= ⋅75
4
∴ x : 1 1⋅ 75
2⋅5
3 ⋅ 75
4
f ( x ) : 2 7 ⋅109 18.125 37 ⋅ 578 68
Trapezoidal integral formula:
b
h
∫ f ( x ) dx = 2 ⎡⎣ y
0
a
4
∴
∫ ( x + x ) dx =
3
1
again I exact
∴
+ yn + 2 ( y1 + y2 + ... + yn −1 )⎤⎦
75
⎡2 + 68 + 2 (7 ⋅109 + 18 ⋅125 + 37 ⋅ 578)⎦⎤ = 73.359 = I approx
2 ⎣
4
⎡ x2 x4 ⎤
⎛ 1 1⎞
= ∫ x + x dx = ⎢ + ⎥ = (8 + 64 ) − ⎜ + ⎟
1
⎝ 2 4⎠
4 ⎦1
⎣2
3
= 72 −
4
= 72 − .75
= 71.25
4
(
error = I exact
⎡cos θ
6. If A (θ ) = ⎢
⎣ sin θ
3
)
− I approx = 2.109.
− sin θ ⎤
then show that
cos θ ⎥⎦
A (θ ) A (φ ) = A (φ ) . A (θ ) = A (θ + φ )
Ans. Refer to the question 2. (vi) 2006.
Solved Question Papers

Q.41
GROUP—C
(Short Answer Type Questions)
Answer any three of the following
3 ¥ 15 = 45
⎡1 2 1 ⎤
⎡2 1 1 ⎤
⎢
⎥
7. (a) If A = ⎢1 −4 1 ⎥ , B = ⎢⎢1 −1 0 ⎥⎥ , show that AB = 6I3. Utilize this result to solve the following
⎢⎣3 0 −3⎥⎦
⎢⎣2 1 −1⎥⎦
system of equations:
2x + y + z = 5
x− y=0
2x + y − z = 1
⎡1 2 1 ⎤
⎡2 1 1 ⎤
Ans. A = ⎢⎢1 −4 1 ⎥⎥ ; B = ⎢⎢1 −1 0 ⎥⎥
⎢⎣3 0 −3⎥⎦
⎢⎣2 1 −1⎥⎦
| det ( B ) = 2 (1 − 0 ) − 1( −1 − 0 ) + 1(1 + 2)
= 2 + 1 + 3 = 6 ≠ 0 ⇒ B −1 exists |
⎡1 2 1 ⎤ ⎡2
AB = ⎢⎢1 −4 1 ⎥⎥ ⎢⎢1
⎢⎣3 0 −3⎥⎦ ⎢⎣2
2 x + y + z = 5⎫
⎪
x − y = 0⎬ ...........(II) Let us
again
2 x + y − z = 0⎪⎭
1 1 ⎤ ⎡6 0 0 ⎤
⎡1 0 0 ⎤
⎥
⎢
⎥
−1 0 ⎥ = ⎢0 6 0 ⎥ = 6 ⎢⎢0 1 0 ⎥⎥ = 6 I 3 (Proved).
⎢⎣ 0 0 1 ⎥⎦
1 −1⎥⎦ ⎢⎣ 0 0 6 ⎥⎦
⎡2 1 1 ⎤
consider P = ⎢⎢1 −1 0 ⎥⎥ (i.e. the coefficient matrix) by given
⎢⎣2 1 −1⎥⎦
problem P = B
Let (II) can be written as PX = Q
⇒ P −1 ( PX ) = P −1 Q (as matrix multiplication is associative)
(
)
⇒ P -1 P X = P −1 Q
⇒ I . X = P −1 Q
⇒ X = P −1 Q ( I → Identity matrix)
= B −1 Q
From (I)
AB = 6 I 3
⇒ ( AB ) B −1 = (6 I 3 ) . B −1
(
)
(
⇒ A BB −1 = 6 I 3 . B −1
⇒ AI = 6. B
−1
)
Q.42

Engineering Mathematics-II
⎡1
⎢6
⎢
1
1
= A= ⎢
⎢6
6
⎢
⎢3
⎣⎢ 6
⇒ A = 6 B −1 ⇒ B −1
⎛ x⎞
∴ X = ⎜ y⎟
⎜ ⎟
⎝ z⎠
⎛1
⎜6
⎜
1
=⎜
⎜6
⎜3
⎜
⎝6
2
6
4
−
6
0
2
6
4
−
6
0
1 ⎤
6 ⎥
⎥
1 ⎥
6 ⎥
⎥
3
− ⎥
6 ⎥⎦
1 ⎞
6 ⎟ ⎛ 5⎞ ⎛ 1⎞
⎟
1 ⎟⎜ ⎟ ⎜ ⎟
0 = 1
6 ⎟⎜ ⎟ ⎜ ⎟
⎝ 1⎠ ⎝ 2⎠
3⎟
− ⎟
6⎠
∴ x = 1, y = 1, z = 2 .
dy
.
(b) Solve: ( y − px )( p − 1) = p and obtain the singular solution. Here p =
dx
Ans. ( y − px )( p − 1) = p
⇒ y − px =
p
p −1
⇒ y − px =
p
→ clairaut’s equation ...........(I)
p −1
differentiating both sides of (I) we get
p = p.1 + x.
⇒x
or,
dp ( p − 1) .1 − p dp
+
⋅
dx
( p − 1)2 dx
dp
dp ⎡
dy ⎤
1
−
Here p = ⎥
2
⎢
dx ( p − 1) dx ⎣
dx ⎦
dp ⎡
1 ⎤
⎢x −
⎥ = 0 From (I)
dx ⎣⎢
( p − 1)2 ⎦⎥
dp
dy
c
= 0 or, p = c .............. (II)
= c ⇒ y = cx +
general
dx
dx
c −1
solution
∴
or,
x=
1
⇒ ( p − 1) =
2
( p − 1)
⇒ p = 1±
2
1
x
1
1
⇒ p −1 = ±
x
x
.........(III)
Solved Question Papers

Q.43
Solving (I) and (III)
⎡
p
1 ⎤
= p ⎢x +
( p − 1) ⎣ ( p − 1) ⎥⎦
⎡
⎤
⎢
⎥
1 ⎞⎢
1
⎛
⎥
= ⎜1 ±
x+
⎟
⎝
1
⎛
⎞⎥
x⎠ ⎢
− 1⎟ ⎥
⎢
⎜⎝1 ±
x ⎠⎦
⎣
y = px +
=
(
y=
(
)×
x ±1
x
(x ± x )
)
2
x ± 1 → singular solution.
(c) Construct the interpolation polynomial for the function y = sin π x, taking the points
1
1
x0 = 0, x1 = , x2 =
6
2
⎛ 1⎞
Hence find f ⎜ ⎟ where y = f ( x ) .
⎝ 3⎠
Ans. y = sin π x
y0 = sin π x0 = sin 0 = 0
y1 = sin
y2 = sin
π
6
π
2
=
1
2
=1
x
x0 = 0
y
y0 = 0
1
6
1
y1 =
2
x1 =
x2 =
1
2
y2 = 1
Since value of x are not equispaced we use Lagrarge’s interpolation to find the polynormial for f(x)
f ( x) =
( x − x0 ) ( x − x2 )
( x − x0 ) ( x − x1 )
( x − x1 )( x − x2 )
y0 +
y1 +
y
( x0 − x1 )( x0 − x2 )
( x1 − x0 ) ( x1 − x2 ) ( x2 − x0 ) ( x2 − x1 ) 2
1⎞ ⎛
1⎞
1
1
⎛
( x − 0) ⎛⎜⎝ x − ⎞⎟⎠
( x − 0) ⎛⎜⎝ x − ⎞⎟⎠
⎜⎝ x − ⎟⎠ ⎜⎝ x − ⎟⎠
1
6
2
2
6
=
×0+
× +
1
1 ⎛ 1⎞
2 ⎛ 1 ⎞ ⎛ 1 1⎞
⎛ 1⎞ ⎛ 1⎞
−
−
−
−
0
−
⎜⎝
⎟⎜
⎟
⎜ ⎟
⎜⎝
⎟⎜
⎟
6⎠ ⎝ 2⎠
6 ⎝ 3⎠
2 ⎠ ⎝ 2 6⎠
1⎞
1⎞
9
⎛
⎛
= −9 x ⎜ x − ⎟ + 6 ( x ) ⎜ x − ⎟ = − x (2 x − 1) + x (6 x − 1)
⎝
⎠
⎝
⎠
2
6
2
= −9 x 2 + 9 x + 6 x 2 − 2
= −3x 2 +
7x
2
Q.44

Engineering Mathematics-II
⎛ 1⎞
⎛ 1⎞ 7 7 1 5
∴ f ⎜ ⎟ = −3 ⎜ ⎟ + = − =
⎝ 3⎠
⎝ 9⎠ 6 6 3 6
8. (a) Solve the differential equation
d2 y
dy
− 5 + 6 y = x 2 e3 x
dx
dx 2
d2 y
dy
− 5 + 6 y = x 2 e3 x
dx
dx 2
We can write the above equation in the standard form as ( D 2 − 5 D + 6) y
d
x 2 e 3 x where D ≡
............(I)
dx
2
Let us consider y = e mx be a trial solution of homogeneous eqn. ( D − 5 D + 6) y = 0
Ans.
∴ Dy = me mx ; D 2 y = m2 e mx
∴ The auxiliary equation is ( m2 e mx − 5me mx + 6e mx ) = 0
∵ e mx ≠ 0 for finite values of x
⇒ m2 − 5m + 6 = 0 ⇒ ( m − 3)( m − 2) = 0
⇒ m = 3, 2
∴ Complementary function = C.F. = C1 e 3 x + c2 e 2 x where c1 , c2 are arbitrary constants now
P.I. =
=
(
1
)
D 2 − 5D + 6
1
e
3x
( D + 3)
2
(x e )
2 3x
+ 5 ( D + 3) + 6
( x2 )
⎡
⎤
1
1
e ax v ( x ) = e ax
v ( x ))⎥
(
⎢∵
f ( D + a)
⎣ f (D)
⎦
(
= e3 x
= e3 x
)
( )
1
x2
D +D
1
x2
D (1 + D )
2
( )
( )
1
−1
. (1 + D ) x 2
D
3x 1
=e
. 1 − D + D 2 − D 3 + ...∞ x 2
D
1 2
= e3 x
[∵ D r x 2 = 0 for r > 2
x − 2x + 2
D
⎛ x3
⎞
= e 3 x ⎜ − x 2 + 2 x ⎟ and D r x 2 = 2! for r = 2]
⎝ 3
⎠
= e3 x
(
(
)( )
)
( )
( )
⎛ x3
⎞
= e3 x ⎜ − x 2 + 2 x⎟
⎝ 3
⎠
⎡ 1
⎤
⎢⎣∵ D ≡ ∫ dx ⎥⎦
Solved Question Papers

Q.45
⎛ x3
⎞
∴ General solution is y = C1 e 3 x + C2 e 2 x + e 3 x ⎜ − x 2 + 2 x ⎟
⎝ 3
⎠
(b) Apply suitable interpolation formula to calculate f (3) correct <??> two significant figures from the following data:
x:
2
4
6
8
10
f(x):
5
10
17
29
49
Ans. Since values of x are equispaced and we are to find f (x) at x were going to apply Newton’s backward
interpolation formula for problem.
Newton’s backward difference table
f ( x ) = yn + s ∇yn +
∇y
∇2y
∇3y
x
y = f (x)
2
5
4
10
5
6
17
7
2
8
29
12
5
3
10
49
20
8
3
∇4y
0
s( s + 1) 2
s( s + 1)( s + 2) 3
∇ yn +
∇ yn + …
2
3!
Here x = 9; h = 2 (difference between values of x); n = 5 we take xn = x5 , yn = y5
x − xn 9 − 10
1
=
= − = −0.5
h
2
2
( −.5)( −.5 + 1)
( −.5)( −.5 + 1)( −.5 + 2)
∴ f (9) = 49 + ( −0.5) 20 +
×3
8+
2
6
1
= 49 − 10 + 4(.5)( −.5) + (.5)(1.5)( −.5)
2
= 39 − 1 − .1875 = 37.8125 ≈ 38
(c) Determine the conditions under which the system of equations
x+ y+ z =1
x + 2y − z = b
∴ xn = 10, yn = 49 ; s =
5 x + 7 y + az = b 2
admits of
(i) only one solution
(ii) no solution
(iii) many solutions.
Ans.
x+ y+ z =1
x + 2y − z = b
5 x + 7 y + az = b 2
Q.46

Engineering Mathematics-II
⎛1 1 1 1 ⎞
⎛1 1 1 ⎞
⎜
⎟
Here the coefficient matrix is P = 1 2 −1 and argumented matrix is = ⎜ 1 2 −1 b ⎟
⎜
⎟
⎜
⎟
⎝ 5 7 a b2 ⎠
⎝5 7 a ⎠
Let us transform Q to Echelon matrix by elementary row operations
1
1 ⎞
⎛1 1 1 1 ⎞
⎛1 1
R2 − R1
⎜ 1 2 −1 b ⎟ ⎯⎯⎯⎯⎯
⎜
→ 0 1 −2
b −1 ⎟
⎜
⎟ R3 − 5 R1 ⎜
⎟
2
⎝5 7 a b ⎠
⎝ 0 2 a − 5 b 2 − 5⎠
1
1
⎛1 1
⎞
⎜
⎯⎯⎯⎯⎯
→ 0 1 −2
b − 1 ⎟ = R (let)
⎟
R3 − 2 R2 ⎜
2
⎝ 0 0 a − 1 b − 2b − 3⎠
1 ⎞
⎛1 1
⎜
∴ P ~ ecelon matrix 0 1 −2 ⎟ = L (say)
⎜
⎟
⎝ 0 0 a − 1⎠
(i) for only one solution Rank of P = Rank of Q = number of unknowns = 3
∴ Rank of L = Rank of R = 3
∴ Rank of L = 3 ⇒ a − 1 ≠ 0 ⇒ a ≠ 1
∴ for a ≠ 1 and for any value of b the gives system has luique solution.
(ii) For no solution Rank of P ≠ Rank of Q ⇒ Rank of L ≠ Rank of R
Let Rank of L = 2 and Rank of R = 3
∴ a − 1 = 0 and b 2 − 2b − 3 ≠ 0
⇒ a = 1 and (b − 3) (b + 1) ≠ 0
⇒ a = 1 and b ≠ −1, 3
∴ the system admits no solution for a = 1, b ≠ −1, 3
(iii) For many solutions Rank of P = Rank of Q < 3 (= no unknowns)
⇒ Rank of L = Rank of R < 3
Since first two rous of R and L are non zero rows
∴ Rank of R = Rank of L < 2
∴ Rank of L = Rank of R = 2
∴ a − 1 = 0 and b 2 − 2b − 3 = 0
⇒ a = 1 and b = −1, 3
∴ when a = 1 ; b = −1, 3 the system admits many solutions.
9. (a) Prove that PT AP is a symmetric or a skew-symmetric matrix according as A is symmetric or skewsymmetric.
Ans. Symmetric q skew-symmetric matrix is always a space matrix
∴ A must be square matrix
Let A be of order n × n and order of P be n × r
∴ AP is of order n × r
Solved Question Papers
but PT is of order r × n
∴ order of PT AP is r × r ⇒ P T AP is a square matrix
[If P is itself a square matrix of order n × n the result is obvious]
∴ ( P T AP )T = ( P T ( AP ))T = ( AP )T ( P T )T '
= ( AP )T P [∵ ( AB)T = BT AT ]
= P T AT P
Now P T AT P = P T AP if AT = A i.e., A is symmetric
= − P T AP if AT = − A i.e., A is skew symmetric
T
T
T
∴ ( P T AP )T = P T AP or ( P AP ) = − P AP according as A is symmetric or skew symmetric.
Hence the result.
⎡ 4 6⎤
(b) Find the eigen values and the eigenvectors of the matrix ⎢
⎥
⎣ 2 9⎦
⎛ 4 6⎞
Ans. Let A = ⎜
⎝ 2 9⎟⎠
For eigen values A − λ I = 0 ⇒
4−λ
6
=0
2
9−λ
⇒ (4 − λ )(9 − λ ) − 12 = 0
⇒ 36 − 13λ + λ 2 − 12 = 0
⇒ λ 2 − 13λ + 24 = 0
∴λ =
13 ± 169 − 96 13 ± 173
=
2
2
13 + 73 13 − 73
,
;
2
2
∴ eigen values are
⎛ x1 ⎞
13 + 73
Let X 1 = ⎜ ⎟ bet the eigen vector corresponding to
x
⎝ 2⎠
2
⎛ 13 + 73 ⎞
⎛ 4 6⎞ ⎛ x1 ⎞ ⎛ 13 + 73 ⎞ ⎛ x1 ⎞
∴ AX 1 = ⎜
=
X1 ⇒ ⎜
⎟
⎟ ⎜⎝ x ⎟⎠
⎝ 2 9⎟⎠ ⎜⎝ x2 ⎟⎠ ⎜⎝
2
2
⎝
⎠
⎠ 2
⇒ 4 x1 + 6 x2 =
⇒ x1 =
=
(
(
12
73 + 5
73 − 5
4
again 2 x1 + 9 x2 =
⎛ 73 + 5 ⎞
13 + 73
x1 ⇒ 6 x2 = ⎜
⎟ x1
2
⎝ 2 ⎠
)x
)
x2
2
13 + 73
x2
2

Q.47
Q.48

⇒ 2 x1 =
Engineering Mathematics-II
(
73 − 5
)x
2
73 − 5
x2
4
⎞
73 − 5 ⎟
when c is a non zero real no.
4 ⎟
1 ⎠
⇒ x1 =
2
⎛
⎞
⎛
73 − 5 ⎟
⎜
c = c⎜
∴ x1 =
⎜ 4
⎟
⎜
⎝
⎝
⎠
c
⎛ y1 ⎞
Let X 2 = ⎜ ⎟ be the eigen vector corresponding to 13 − 73
⎝ y2 ⎠
2
⎛ 13 − 73 ⎞
⎛ 4 6⎞ ⎛ y1 ⎞ ⎛ 13 − 73 ⎞ ⎛ y1 ⎞
∴ AX 2 = ⎜
X2 ⇒ ⎜
=
⎟
⎟ ⎜⎝ y ⎟⎠
⎝ 2 9⎟⎠ ⎜⎝ y2 ⎟⎠ ⎜⎝
2
2
⎝
⎠
⎠ 2
∴ 4 y1 + 6 y2 =
2 y1 + 9 y2 =
⎛ 73 − 5 ⎞
13 − 73
73 + 5
y1 ⇒ ⎜
y2
⎟ y1 = −6 y2 ⇒ y1 = −
2
4
⎝ 2 ⎠
13 − 73
5 + 73
1
y2 ⇒ −2 y1 =
y2 ⇒ 1 y1 = −
2
2
4
(
⎡
⎤
1
⎢
∴ X 2 = k ⎢ − 5 + 3 ⎥⎥ where k ≠ 0 real number.
4
⎥⎦
1
⎣⎢
(
)
(c) Solve by Cramer’s rule the following system of equations :
3x + y + z = 4
x − y + 2z = 6
x + 2 y − z = −3
Ans. 3 x + y + z = 4
x − y + 2z = 6
x + 2 y − z = −3
3 1 1
Let Δ = 1 −1 2 = 3(1 − 4) − 1( −1 − 2) + 1(2 + 1)
1 2 −1
= −9 + 3 + 3 = −3 ≠ 0
4 1 1
Δ1 = 6 −1 2 = 4(1 − 4) − 1( −6 + 6) + 1(12 − 3)
−3 2 −1
= −12 + 9 = −3
3 4 1
Δ 2 = 1 6 2 = 3( −6 + 6) − 4( −1 − 2) + 1( −3 − 6)
1 −3 −1
= 12 − 9 = 3
)
73 + 5 y2
Solved Question Papers
3 1 4
Δ 3 = 1 −1 6 = 3(3 − 12) − 1( −3 − 6) + 4(2 + 1) = −27 + 9 + 12 = 6
1 2 −3
∴x =
Δ
Δ1
Δ
= 1; y = 2 = −1; z = 3 = 2
Δ
Δ
Δ
10. (a) What is meant by linear independence of a set of n-vectors ?
Ans. Refer to the soln of Q. 13. (a)/2008.
(b) Solve by the method of variation of parameters the equation
d2 y
+ 9 y = sec 3 x.
dx 2
d2 y
+ 9 y = sec 3 x
dx 2
We can write the given equation in standard form as ( D 2 + 9) y = sec 3 x
Ans.
⎡
d ⎤
⎢⎣ D = dx ⎥⎦ .
Let, y = e mx be the trial solution of homogereous eqution ( D 2 + 9) y = 0
∴ Dy = me mx ; D 2 y = m2 e mx
∴ auxiliary equation is ( m2 + 9) me mx ≠ 0
∵ e mx ≠ for finite values of x
∴ m2 + 9 = 0 ⇒ m = ± 3i where i = −1
∴ C.F.: c1 cos 3 x + c2 sin 3 x ; (c1 , c2 are arbitrary constants)
Let us consider P.I. = u cos 3 x + v sin 3 x where u, v are function of x .........(1)
∴ y p = u cos3 x + v sin 3 x
Dy p = −3u sin 3 x + 3v cos3 x + cos3 x
Let
cos3x
du
dv
+ sin
=0
dx
dx
du
dv
+ sin 3 x
dx
dx
..........(II)
∴ D 3 y p = −9u cos3 x + ( −3sin 3 x )
du
⎛ dv ⎞
− 9v sin 3 x + 3cos3 x ⎜ ⎟ ........(III)
⎝ dx ⎠
dx
substitution (II) and (III) in (I) we get
∴ 3cos3 x
dv
du
− 3sin 3 x
= sec3 x ........... (IV)
dx
dx
0
sin 3 x
sin 3 x 3cos3 x
du
− tan 3 x
1
∴
=
=
= − tan 3 x
cos3 x
sin 3 x
dx
3
3
−3sin 3 x 3cos3 x

Q.49
Q.50

Engineering Mathematics-II
1
1
1
tan 3 xdx = − logsec3 x = log cos3 x
∫
3
9
9
cos3x
0
−3sin 3 x sec3 x
dv
1
=
=
cos3 x
sin 3 x
dx
3
−3sin 3 x 3sec3 x
u=−
1
x
∴ ∫ dv = ∫ dx = v =
3
3
x
⎛1
⎞
∴ P.I. = ⎜ log cos3 x ⎟ cos3 x + (sin 3 x )
⎝9
⎠
3
1
x
∴ general solution y = c1 cos3 x + c2 sin 3 x + cos3 x log cos3 x + sin 3 x.
9
3
(b + c)2
a2
a2
(c) Prove that Δ = b 2
( c + a) 2
b2
= 2abc ( a + b + c)3
c2
c2
( a + b) 2
Ans. Refer to the solution of Q. 2(a) (2nd part)/2004.
Index
A
Acyclic graph 2.44
Adjacency matrix 2.29
Application of Laplace Transform (to) 7.1
Differential equation with 7.1
Variable coefficients 7.10
Difference equation 7.16
Electric circuits 7.6
Evaluation of integrals 7.29
Mechanical system 7.9
Ordinary differential equation 7.10
Partial differential equation 7.25
Simultaneous differential equations 7.13
Applications of linear differential equation 1.20
Auxiliary equation 1.39
B
Bernoulli’s equation 1.4
Beta function 4.1
Bipartite graph 2.4
Binary tree 2.51
Bridge 2.20
C
Cauchy–Euler homogeneous linear equation 1.61
Cayley’s formula 2.56
Chromatic number of graph 2.37
Circuit 2.11
Colouring of graph 2.37
Complete graph 2.3
Complementary function 1.39
Complement of a subgraph 2.7
Connected graph 2.5
Cut edge 2.20
Cut set 2.68
Cycle graph 2.5
Cycle of length k 2.12
Convolution theorem for Laplace transform 6.16
D
Degree of a vertex 2.41
Diameter of a connected graph 2.68
Dijkstra’s algorithm 2.58
Differential equation 1.1
Degree of 1.1
Exact 1.15
General solution of 1.14
Linear 1.12
Order of 1.1
Particular solution of 1.4
Partial 1.1
Solution of 1.3
With variable coefficients 1.54
Directed graph 2.41
Dirichlet’s theorem 4.11
Duplication formula 4.7
E
Edge 2.1
Equation reducible to exact equation 1.18
Equation reducible to homogeneous form 1.11
Equation reducible to linear differential equation 1.14
Euler’s circuit 2.18
Euler’s formula for planar graph 2.34
Euler’s path 2.19
Euler’s Theorem 2.3
Euler’s reflection formula 4.7
Eulerian graph 2.15
F
First Theorem of Graph Theory 2.3
Four Colour Theorem 2.41
Fourier–Bessel expansion of a continuous function 1.88
Fourier–Legendre expansion 1.95
Fully binary tree 2.51
G
Gamma function 4.5
Graph 2.1
Greedy algorithm 2.63
H
Hamiltonian circuit 2.21
Hamiltonian graph 2.21
Harmonic oscillator 1.26
Heavyside’s expansion formule 6.2
Homogeneous equations 1.8
I
Incidence matrix of a graph 2.30
Isomorphism of graphs 2.9
Integrating factor 1.13
J
Jacobi’s series 1.88
I.2
„ Index
K
Neumann function 1.77
Newton law of cooling 1.28
Kruskal’s algorithm 2.65
Kuratowski’s graph 2.35
Kuratowski’s subgraph 2.36
Kuratowski’s theorem 2.35
O
L
P
Laplace transform (of)
Bessels’ function of order zero 5.11
Derivative 5.14
Dirac-delta function 5.7
Error function 5.4
Full wave rectified sine wave 5.21
Half wave rectified sinusoidal 5.21
Heavyside’s unit step function 5.25
Integrals 5.19
Periodic function 5.20
Pulse of unit height 5.6
Rectangular function 5.7
Sawtooth wave 5.22
Square-wave function 5.9
Triangular pulse 5.6
Triangular wave 5.22
Unit ramp function 5.2
Unit step function 5.2
Legendre’s equation 1.89
Legendre’s linear equation 1.64
Leibnitz’s linear equation 1.12
Lerch’s theorem 6.1
Limiting theorem for Laplace transform 5.22
Linear differential equation with constant coefficients 1.37
Linear homogeneous partial differential equation with
constant coefficients 1.39
Lipschitz condition 1.4
Liouville’s theorem 4.11
Particular integral 1.39
Path matrix 2.42
Peano’s existence theorem 1.5
Picard’s existence and uniqueness theorem 1.4
Planar graph 2.30
Polish form 2.73
Postfix form 2.75
Prefix form 2.73
Prim algorithm 2.63
M
Mass-spring system 7.9
Matrix representation of graph 2.29
Max-flow, Min-cut Theorem 2.80
Method of variation of parameters 1.51
to find particular integral 1.51
Method of solution by changing independent
variable 1.55
Method of solution by changing dependent variable 1.55
Method of undetermined coefficients 1.57
Method of reduction of order 1.59
Minimal spanning tree 2.63
Modeling 1.1
Moore BFS algorithm 2.62
N
n-ary tree 2.51
Nearest neighbour of a vertex 2.27
Orthogonal trajectories 1.34
R
Rate problems 1.31
Reachability matrix 2.42
Region of a planar graph 2.31
Regular graph 2.3
Reverse polish form 2.75
Rodrigue’s formula 1.39
S
Searching of a tree 2.72
Secular term 7.2
Separable differential equation 1.5
Seven bridges problem 2.19
Simple pendulum 1.70
Simultaneous linear differential equation 1.65
Solution in series 1.71
Spanning tree 2.65
Shortest path problem 2.75
Standard cases of particular integral 1.43
Strongly connected digraph 2.42
T
Tautochrone curve 7.23
Terminal vertex 2.41
Traveling salesperson problem 2.26
Tree 2.44
V
Vertex colouring 2.37
W
Walks 2.11
Warshall’s algorithm 2.42
Weight of an edge 2.26
Weighted graph 2.26
Welsh–Powel algorithm 2.38
Wronskian 1.51